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I would like you to estimate the population proportion of adults without children that visit the local zoo. 500 total, children included, visit the zoo each day. Out of that only 120 are adults without children. Calculate a 95% confidence interval for the population proportion. Assume a random sample. Thank you so much.

Hello, I need help with constructing a P chart that contains 20 days and 5 stores. In the calculating step, do I have to compute number of B1 through B5 stores and divide the number by 20 to arrive in the proportion or do I have to the B1 through B5 stores and divide the number by 20? Please help with this question and show computations. Thank you!

 

As part of a larger project to study the behavior of stressed-skin panels, a structural component being used extensively in North America, an article reported on various mechanical properties of Scotch pine lumber specimens. Data on the modulus of elasticity (MPa) obtained 1 minute after loading in a certain configuration and 4 weeks after loading for the same lumber specimens is presented here.

  Observation       1 min   4 weeks     Difference   1             10,450 9,130   1320 2             16,620 13,250   3370 3             17,500 14,720   2780 4             15,480 12,710   2770 5             12,960 10,120   2840 6             17,260 14,570   2690 7             13,400 11,220   2180 8             13,900 11,100   2800 9             13,630 11,420   2210 10             13,210 10,940   2270 11             14,370 12,110   2260 12             11,700 8,640   3060 13             15,470 12,590   2880 14             17,840 15,090   2750 15             14,060 10,550   3510 16             14,760 12,230   2530

Calculate an upper confidence bound for the true average difference between 1-minute modulus and 4-week modulus; first check the plausibility of any necessary assumptions. (Use 𝛼 = 0.05. Round your answer to two decimal places.)
 MPa

Interpret your upper confidence bound.

We are 95% confident that the true mean difference between 1-minute modulus and 4-week modulus is never the upper bound. We are 95% confident that the true mean difference between 1-minute modulus and 4-week modulus is at least the upper bound.   We are 95% confident that the true mean difference between 1-minute modulus and 4-week modulus is at most the upper bound. We are 95% confident that the true mean difference between 1-minute modulus and 4-week modulus is exactly the upper bound.

A simple random sample of 70 items resulted in a sample mean of 90. The population standard deviation is o 10. (a) Compute the 95% confidence interval for the population mean. (Round your answers to two decimal places.) to (b) Assume that the same sample mean was obtained from a sample of 140 items. Provide a 95% confidence interval for the population mean. (Round your answers to two decimal places.) to (c) What is the effect of a larger sample size on the interval estimate? A larger sample size does not change the margin of error. A larger sample size provides a larger margin of error. A larger sample size provides a smaller margin of error.

 

Find the margin of error for a 95% confidence interval for estimating the population mean when the sample standard deviation equals 90, with a sample size of) 324 and () 1444. What is the effect of the sample size? () Find the margin of error for a 95% confidence interval for estimating the population mean when the sample standard deviation equals 90 with a sample size of 324 (Round to two decimal places as needed.) (1) Find the margin of error for a 95% confidence interval for estimating the population mean when the sample standard deviation equals 90 with a sample size of 1444 (Round to two decimal places as needed.) What is the effect of the sample size on the margin of error? Choose the correct answer below. OA The margin of error is the same for a sample sizes OB. The larger sample size decreases the margin of error, O C. The larger sample size increases the margin of error, OD. There is no correlation between the sample size and the margin of error

A professional baseball pitcher takes 14.61 seconds to throw each pitch, on average. Assume the pitcher's times per pitch follow the normal probability distribution with a standard deviation of 2.8 seconds. Complete parts a through d. a. What is the probability that a random sample of 10 pitches from this pitcher will have a mean less than 14 seconds? P(x< 14) = (Round to four decimal places as needed.) b. What is the probability that a random sample of 30 pitches from this pitcher will have a mean less than 14 seconds? PX< 14) = (Round to four decimal places as needed.) c. What is the probability that a random sample of 50 pitches from this pitcher will have a mean less than 14 seconds? PX< 14) = (Round to four decimal places as needed.) d. Explain the difference in these probabilities. Choose the correct answer below. O A. With a larger sample size, the standard error of the mean decreases and the sample means tend to move closer to the population mean of 14.61 seconds. Therefore, the probability of observing a sample mean less than 14 seconds decreases as the sample size increases. OB. With a larger sample size, the standard error of the mean stays the same and the sample means stay the same. Therefore, the probability of observing a sample mean less than 14 seconds decreases as the sample size increases. O c. With a larger sample size, the standard error of the mean increases and the sample means tend to move closer to the population mean of 14.61 seconds. Therefore, the probability of observing a sample mean less than 14 seconds decreases as the sample size increases. OD. With a larger sample size, the standard error of the mean increases and the sample means tend to move further away from the population mean of 14.61

A study found that highway drivers in one stare traveled at an average speed of 60.8 miles per hour (MPH). Assume the population standard deviation is 6.7 MPH. Complete parts a through d below. a. What is the probability that a sample of 30 of the drivers will have a sample mean less than 59 MPH? P (*<59) - (Round to four decimal places as needed.) b. What is the probability that a sample of 50 of the drivers will have a sample mean less than 59 MPH? P(x<59) - (Round to four decimal places as needed.) c. What is the probability that a sample of 70 of the drivers will have a sample mean less than 59 MPH? P(x<50) - (Round to four decimal places as needed.) d. Explain the difference in these probabilities and the same means the population mean of 60.8 MPH. Therefore, the probability of observing a As the sample size increases the standard error of the mean sample mean less than 50 MPH According to a research Institution, men spent an average of 136 87 on Valentine's Day men who celebrate Valentine's Day was selected. Complete parts a through is in 2009. Assume the standard devon for this population is $45 and that it is normally distrutod. A random sample of 10 a. Calculate the standard error of the mean (Pound to two decimal places as needed.) b. What is the probability that the sample mean will be less than $1257 PG<$125) - (Round to four decimal places as needed.) c. What is the probability that the sample mean will be more than $145? P(x>$145) = (Round to four decimal places as needed.) d. What is the probability that the sample mean will be between $110 and $1607 P ($110 SXS $160) (Round to four decimal places as needed) e. Identify the symmetrical interval that includes 95% of the sample means the true population mean is $136.87 $ sxss (Round to the nearest dollar as needed)

The population mean and standard deviation are given below. Find the required probability and determine whether the given sample mean would be considered unusual. For a sample of n=63, find the probability of a sample mean being less than 23.5 if μ=24 and σ=1.34. For a sample of n=63 ,the probability of a sample mean being less than 23.5 if μ=24 and σ=1.34 is ______________ (round to four decimal places).

Suppose that we want to estimate the number of holding penalties assessed during a college football game. The sample of games we pick has a mean of 2.9 penalties per game and a standard deviation of 0.7 penalties per game. For each of the following sampling scenarios, determine which test statistic is appropriate to use when making inference statements about the population mean. (In the table, Z refers to a variable having a standard normal distribution, and t refers to a variable having a t distribution.) Sampling Scenario could use either Z or unclear O (1) The sample has size 15, and it is from a population with a distribution about which we know very little (2) The sample has size 75, and it is from a non- normally distributed population with a known standard deviation of 0.65. (3) The sample has size 16, and it is from a normally distributed population with a known standard deviation of 0.65. | 0 0 0 0 (4) The sample has size 90, and it is from a non- normally distributed population. 0 0 (5) The sample has size 10, and it is from a normally distributed population with unknown standard deviation. 0 0 o o xs ?

A sample of 44 observations is selected from one population with a population standard deviation of 3.1. The sample mean is 101.0. A sample of 56 observations is selected from a second population with a population standard deviation of 5.0. The sample mean is 99.5. Conduct the following test of hypothesis using the 0.10 significance level. : = : ≠

1) Scores from a questionnaire measuring anxiety form a normal distribution with a population mean of μ = 50 and a population standard deviation of σ = 10. What is the z for obtaining a sample mean greater than M = 53 for the following sample size n = 5 2) Scores from a questionnaire measuring anxiety form a normal distribution with a population mean of μ = 50 and a population standard deviation of σ = 10. What is the z for obtaining a sample mean greater than M = 53 for the following sample size n = 17 3) Scores from a questionnaire measuring anxiety form a normal distribution with a population mean of μ = 50 and a population standard deviation of σ = 10. What is the z for obtaining a sample mean greater than M = 53 for the following sample size n = 42

Consider the following hypotheses: : = 4,500 : ≠ 4,500 The population is normally distributed with a population standard deviation of 700. Compute the value of the test statistic and the resulting -value for each of the following sample results. For each sample, determine if you can "reject/do not reject" the null hypothesis at the 10% significance level. z table t table

(1 point) The contents of 33 cans of Coke have a mean of x⎯⎯⎯=12.15. Assume the contents of cans of Coke have a normal distribution with standard deviation of σ=0.1. Find the value of the test statistic z under the null hypothesis that the population mean is μ=12. The test statistic is

(b) (9 points) Find a 95% confidence interval for the population mean w. Do this by completing the following: (1) In this case, a = • Zal2 = (2) Za/2 (3) The confidence interval is (c)(3 points) Interpret your result: 7. (15 points, 3 points each) Given the sample mean of a simple random sample the sample size, the sample standard deviation, the test hypothesis and significance level: * = 16,5 = 3, n = 22,H: = 13, H.: u < 13, 5% significand level. Perform the one mean t-test. (a) The value of the test statistic t = (b) The test statistic has df = (c) The critical value is (d) Reject or not reject the null hypothesis (e) Interpretation:

An experiment is planned to test the hypothesis Ho:p=8.0 versus Hai>8.0. What sample size is required to reject Ho with 90% power when u=8.2? The process is known to be normally distributed with standard deviation Ox=0.2. (Note: The relevant formula is available on page seven of the formulae pack) (3 marks) Quality & Reliability 2 (QURL4111): Formulae Pack Page 8 of 13 Sample Standard Deviation (Estimator of o) (x - x) n-1 1-Sample 2 Test: Test Statistic 2-X-u 1-Sample T test: Test Statistic X-u which has - which has at distribution with n-1 degrees of freedom 2-Sample Z test: Test Statistic X.-X Z= vnn 2-Sample t test: Test Statistic X-Y t= whic which has at distribution with n,+ny-2 degrees of freedom s1 + S yn 17. where S- (- 1)Sun - 1)S n +n - 2

For (a) and (b), is the statement a null hypothesis or an alternative hypothesis? a. The proportion of adults who favor legalized gambling equals 0.50. b. The proportion of all college students who are regular smokers is less than 0.27, the value it was ten years ago. c. Introducing notation for a parameter, state the hypotheses in (a) and (b) in terms of the parameter values. a. Is the statement in part (a) the null or alternative hypothesis? Altenative hypothesis O Null hypothesis b. Is the statement in part (b) the null or altermative hypothesis? Altemative hypothesis O Null hypothesis c. Choose the correct null and altemative hypotheses for part (a). O A. OB. Oc. Ho: P>0.50 Ho: P0.50 Ho: P=0.50 H: p=0.50 Hp=0.50 H:p<0.50 OD. OE. OF. H:p<0.50 Ho: P=0.50 Hg: p=0.50 H: p=0.50 H:p-0.50 H:Pr0.50 Choose the correct null and altenative hypotheses for part (b). OA. OB. OC. Hạ: p0.27 Ho: p=0.27 Ho: p>0.27 H:P>0.27 H:p=0.27 LZ0 >d;"H OD. OE. OF. H: p<0.27 Hp=0.27 H: pH0.27 Họ: P=027 H:p=0.27 H: p4027

Use the following scenario for questions 6 – 8. In 2016-17, the average SAT score of the UVA first year class was 1420 out of 1600. An advisor in the statistics department is interested in determining whether statistics majors from that class have higher SAT scores than the average. He tests H,:µ = 1420 against Hạ: µ > 1420, where u is the mean SAT score for statistics majors from the 2016-17 first year class. In a simple random sample of 30 statistics majors, he finds a sample mean SAT score of 1440 with sample standard deviation of 40.

6. What is the value of the test statistic for this significance test?

a. 0.09

b. 0.5

c. 2.73

d. 20

7. From what distribution is this test statistic?

a. N(0, 1)

b. N(1420, 7.30)

c. t distribution with 29 degrees of freedom

d. t distribution with 28 degrees of freedom

8.) The p-value for this test is 0.005. Which of the following is a proper interpretation of this p- value?

a.There is a 0.5% chance that the null hypothesis is true.

b. If the population mean SAT score really is 1420, there's a 0.5% chance that the sample a. mean SAT score from 30 statistics majors would equal 1440.

c.If the population mean SAT score really is 1420, there's a 0.5% chance that the sample mean SAT score from 30 statistics majors would be 1440 or lower.

d. If the population mean SAT score really is 1420, there's a 0.5% chance that the sample mean SAT score from 30 statistics majors would be 1440 or higher.

(a) In testing hypotheses, which of the following would be strong evidence against the null hypothesis? 

A. Obtaining data with a small P-value. 

B. Obtaining data with a large P-value.

C. Using a small level of line significance.

D. Using a a large level of significance.

(b) The P-value of  a test of a null hypothesis is 

 

A. The probability of the null hypothesis is true.

B. The probability, assuming the null hypothesis is false, that the test statistic will take a value at least as extreme as that actually observed.

C. The probability, assuming the null hypothesis is true, that the test statistic will take a value at least as extreme as that actually observed.

D. The probability of the null hypothesis is false.

(c) In formulating hypotheses for a statistical test of significance, the null hypothesis is often 

A. 0.05 

B. the probability of observing the data you actually obtained 

C. a statement of ''no effect'' or '' no difference''.

D. a statement that the data are 0.

3. A researcher claimed that less than 20% of adults smoke cigarettes. y selected adults showed that 17% of the respondents smoke. Is this evidence researcher's claim? adults smoke cigarettes. A Gallup survey of 1016 (a) What is the sample proportion, p, for this problem? (b) State the null and alternative hypothesis. Ho: p = H :p Note that you should have written the same numb uld have written the same number in the null and alternative hypotheses. This is the value that the researcher is claiming. The sample proportion should never goth null or alternative hypotheses. Therefore, the num ative hypotheses. Therefore, the number you have in (a) should not be in the null or alternative hypothesis. (c) Calculate the test statistic using the following formula: pg where p is the value you stated in the null hypothesis and g is 1-P. " hypothesis and g is 1-2. (a) now, using either your calculator or the normal distribution table, look up the area that cor- l'esponds to the 2-score you calculated in (c). You don't need to subtract from 1 because the alternative hypothesis has < in it. This value is called the p-value. p-value= The p-value represents the probability of getting a p of .17 or smaller if the null hypothesis is true (p= 20). Therefore, if this probability is small then we would doubt that the null hy- pothesis is true (i.e. if it is not very likely to get p if the true proportion is 20% then this gives us reason to doubt that p = 2). Thus, small p-values support the alternative hypothesis, P-values less than 10 or .05 are generally considered small. (e) Circle the correct answer: i. The p-value was small which gives us evidence that the actual percent of adults who smoke is less than 20%. ii. The p-value was large so we do not have evidence that the actual percent of adults who smoke is less than 20%. The answer your circled in (e) is the conclusion to the hypothesis test. We always state conclusions in context of the problem and we state whether we did or did not have evidence for the alternative hypothesis. In this case, we did have evidence for the alternative hypothesis that the percentage of adult smokers is less than 20%

sers experienced 1. The drug Ziac is used to treat hypertension. In a clincial test, 3.2% of 221 Ziac users experie dizziness. Calculate and interpret a 95% confidence interval for the proportion of Ziac users who experience dizziness. 2. In a study designed to test the effectiveness of echinacea for treating upper respiratory tract infectione in children, 337 children were treated with echinacea and 370 other children were given a placeho The number of days of peak severity of symptoms for the echinacea group had a mean of 6.0 days and a standard deviation of 2.3 days. The numbers of days of peak severity of symptoms for the placebo group had a mean of 6.1 days and a standard deviation of 2,4 days. (a) Construct the 95% confidence interval for the mean number of days of peak severity of symp- toms for those who receive echinacea treatment. Use ta/2 = 1.969. Interpret the confidence interval. (b) Construct the 95% confidence interval for the mean number of days of peak severity of symp- toms for those who are given a placebo. Use ta/2 = 1.969. Interpret the confidence interval. (c) Compare the two confidence intervals. What do the results suggest about the effectiveness of echinacea? Justify your answer based on the confidence intervals. 3. A researcher claimed that less than 20% of adults smoke cigarettes. y selected adults showed that 17% of the respondents smoke. Is this evidence researcher's claim? adults smoke cigarettes. A Gallup survey of 1016 (a) What is the sample proportion, p, for this problem? (b) State the null and alternative hypothesis. Ho: p = H :p Note that you should have written the same numb uld have written the same number in the null and alternative hypotheses. This is the value that the researcher is claiming. The sample proportion should never goth null or alternative hypotheses. Therefore, the num ative hypotheses. Therefore, the number you have in (a) should not be in the null or alternative hypothesis. (c) Calculate the test statistic using the following formula: pg where p is the value you stated in the null hypothesis and g is 1-P. " hypothesis and g is 1-2. (a) now, using either your calculator or the normal distribution table, look up the area that cor- l'esponds to the 2-score you calculated in (c). You don't need to subtract from 1 because the alternative hypothesis has < in it. This value is called the p-value. p-value= The p-value represents the probability of getting a p of .17 or smaller if the null hypothesis is true (p= 20). Therefore, if this probability is small then we would doubt that the null hy- pothesis is true (i.e. if it is not very likely to get p if the true proportion is 20% then this gives us reason to doubt that p = 2). Thus, small p-values support the alternative hypothesis, P-values less than 10 or .05 are generally considered small. (e) Circle the correct answer: i. The p-value was small which gives us evidence that the actual percent of adults who smoke is less than 20%. ii. The p-value was large so we do not have evidence that the actual percent of adults who smoke is less than 20%. The answer your circled in (e) is the conclusion to the hypothesis test. We always state conclusions in context of the problem and we state whether we did or did not have evidence for the alternative hypothesis. In this case, we did have evidence for the alternative hypothesis that the percentage of adult smokers is less than 20%

Vetermining null and alternative hypotheses A college professor states that this year's entering students appear to be smarter than entering students from previous years. The college's records indicate that the mean 10 for entering students from earlier years is 110. Suppose that we want to sample a small number of this year's entering students and carry out a hypothesis test to see if the professor's statement can be supported. State the null hypothesis H. and the alternative hypothesis H, that we would use for this test.

A presidential candidate's aide estimates that, among all college students, the proportion p who intend to vote in the upcoming election is at most 65%. IF 182 out of a random sample of 255 college students expressed an intent to vote, can the aide's estimate be rejected at the 0.05 level of significance? Perform a one-tailed test. Then fill in the table below. Carry your intermediate computations to at least three decimal places and round your answers as specified in the table, The null hypothesis: > The alternative hypothesis 10 D-a Oso020 The type of test statistics (Choose one) DDDDDD The value of the test statistic (Round to at least three decimal places) The critical value at the 0.05 bevel of significance Round to at least three decimal places) Can weet the desestimate the oportion of students when to vote is most 65

A psychologist specializing in marriage counseling claims that, among all married couples, the proportion p for whom her communication program can prevent divorce is at least 76%. In a random sample of 230 married couples who completed her program, 160 of them stayed together. Based on this sample, can we reject the psychologist's claim at the 0.05 level of significance? Perform a one-tailed test. Then fill in the table below Carry your intermediate computations to at least three decimal places and round your answers as specified in the table. The nut hypothesis OP The witerative hypothesis: 40 The type of test statistic D-a Oso DDDDD The value of the test statistics (Round to at least three decimal places) X The critical value 0.05 level of significance (Round to at least three Decimal places) Can we reject the psychologists aim that the proportion of and coules for women pull i Wuuston Attempt 1 of 1 Timo Remaining: 1:23:42 . Penormone call to low Carry your intermediate computations to at least three decimal places and round your answers as specified in the table. The null hypothesis: HOP The alternative hypothesis: The type of test statistic: (Choose one) The value of the test statistic: (Round to at least three decimal places.) 0-0 Oso 020 0-0 DO 0> X 5 ? The critical value at the 0.05 level of significance: (Round to at least three decimal places.) Can we reject the psychologist's daim that the proportion of married couples for whom her program can prevent divorce is at least 76%? Yes No Continue

A psychologist specializing in marriage counseling daims that, among all married couples, the proportion p for whom her communication program.com divorce is at least 79%. In a random sample of 210 married couples who completed her program, 163 of them stayed together. Based on this sample reject the psychologist's daim at the 0.05 level of significance? Perform a one-tailed test. Then fill in the table below. Carry your intermediate computations to at least three decimal places and round your answers as specified in the table. The null hypothesis: u OP SA The alternative hypothesis: The type of test statistic: (Choose one) 0-0 DO 020 0-0 oa DB The value of the test statistic: (Round to at least three decimal places Save For Later Continue Type here to search The null hypothesis: The alternative hypothesis: H :1 The type of test statistic: (Choose one) 0=o oso 020 The value of the test statistic: (Round to at least three decimal places.) 0-0 Oca X 5 0 ? The critical value at the 0.05 level of significance (Round to at least three decimal places.) Can we reject the psychologist's claim that the proportion of married couples for whom her program can prevent divorce is at least 79°?

A decade-old study found that the proportion of high school seniors who felt that getting rich was an important personal goal was 68%. Suppose that we have reason to believe that this proportion has changed, and we wish to carry out a hypothesis test to see if our belief can be supported. State the null hypothesis Ho and the alternative hypothesis H, that we would use for this test. H0 1,0 D-0-0 x 5 ?

wuestion 11 of 12 point Question Attempt 1 of 1 Time Remaining 31:32 A decade-old study found that the proportion, p. of high school seniors who believed that 'getting rich' was an important personal goal was 70%. A researcher decides to test whether or not that percentage still stands. He finds that, among the 215 high school seniors in his random sample, 142 believe that getting rich' is an important goal. Can he conclude, at the 0.1 level of significance, that the proportion has indeed changed? Perform a two-tailed test. Then fill in the table below. Carry your intermediate computations to at least three decimal places and round your answers as specified in the table. The null hypothesis: HOP The alternative hypothesis: The type of test statistic (Choose one DHO DO D The value of the test statistic (Round to at least three decimal places) The p-value: (Round to at least three decimal places.) Can we conclude that the proportion of high school seniors who believe that getting rich is an Question 11 of 12 (1 point) | Question Attempt 1 of 1 Time Remaining Carry your intermediate computations to at least three decimal places and round your answers as specified in the tabl The null hypothesis: The alternative hypothesis: The type of test statistic: O=DOso 020 (Choose one) D-OOO 00 The value of the test statistic: (Round to at least three decimal places.) The p-value: (Round to at least three decimal places.) Can we conclude that the proportion of high school seniors who believe that getting rich' is an important goal has changed? Yes No

A decade old study found that the proportion, p, of high school seniors who believed that 'getting rich' was an important personal goal was 70%. A researcher decides to test whether or not that percentage still stands. He finds that, among the 215 high school seniors in his random sample, 142 believe that 'getting rich' is an important goal. Can he conclude, at the 0.1 level of significance, that the proportion has indeed changed? Perform a two-tailed test. Then fill in the table below. Carry your intermediate computations to at least three decimal places and round your answers as specified in the table. H,:0 The null hypothesis: The alternative hypothesis: The type of test statistic: D-0 OSO (Choose one) The value of the test statistic: (Round to at least three decimal places.) The p-value: LCReend to at least three Continue Submit Assignment olo as speditied in the table. H, :0 The null hypothesis: The alternative hypothesis: The type of test statistic: (Choose one) D=0 OSO O0 The value of the test statistic: (Round to at least three decimal places.) ? The p-value: (Round to at least three decimal places.) Can we conclude that the proportion of high school seniors who believe that 'getting rich' is an important goal has changed? O Yes ONo 미미

OLUTIP The accompanying table provides the data for 100 room inspections at each of 25 hotels in a major chain. Management would like the proportion of nonconforming rooms to be less than 2%. Formulate a one-sample hypothesis test for a proportion and perform the calculations using the correct formulas and Excel functions. Use a level of significance of 0.05. .: Click the icon to view the room inspection data. Is there sufficient evidence at the 0.05 level of significance that the proportion of nonconforming rooms to be less than 2%? Determine the null hypothesis, Ho, and the alternative hypothesis, H. (Type integers or decimals. Do not round.) Compute the test statistic. (Round to two decimal places as needed.) Find the p-value for the test. (Round to three decimal places as needed.) State the conclusion The p-value is the chosen value of a, so the null hypothesis. There is evidence to conclude that the proportion of nonconforming rooms will be less than 2%. Sample Room Inspection Results Rooms Nonconforming Inspected Rooms 100 100 100 Fraction Nonconforming 0.04 0.02 0.00 0.00 0.02 0.06 0.04 0.07 100 100 0.02 100 100 0.04 0.01 0.03 0.01 0.03 100 0.05 100 100 0.02 0.02 0.02 0.05 0.02 0.03 0.05 0.01 100 0.01 0.01

A financial advisor believes that the proportion of investors who are risk-averse (that is, try to avoid risk in their investment decisions) is at least 0.6. A survey of 30 investors found that 18 of them were risk-averse. Formulate a one-sample hypothesis test for a proportion to test this belief. Is there sufficient evidence at the 0.05 level of significance that the proportion of investors who are risk-averse is at least 0.6? Determine the null hypothesis, Ho, and the alternative hypothesis, H, Ho O (Type integers or decimals. Do not round.) Compute the test statistic. (Round to two decimal places as needed.) Find the p-value for the test. (Round to three decimal places as needed.) State the conclusion The p-value is the chosen value of a, so the null hypothesis. There is evidence to conclude that the proportion of investors who are risk-averse is at least 0.6.

Question Help The price of a certain combo meal at different franchises of a national fast food company varies from $5.00 to $17.36 and has a known standard deviation of $2.29. A sample of 26 students in an online course that includes students across the country stated that their average price is $6.00. The students have also stated that they are generally unwilling to pay more than $6.75 for this meal. Formulate and conduct a hypothesis test to determine if you can conclude that the population mean is less than $6.75. Use a level of significance of 0.01. Is there sufficient evidence at the 0.01 level of significance that the population mean is less than $6.75? Determine the null hypothesis, Ho, and the alternative hypothesis, H. (Type integers or decimals. Do not round.) Compute the test statistic. (Round to two decimal places as needed.) Find the p-value for the test. (Round to three decimal places as needed.) State the conclusion The p-value is the chosen value of a, so the null hypothesis. There is evidence to conclude that the population mean is less than $6.75

Utwo como Test wh a t the correo Cardacordebout a y the land hypothes for на ОА HI ор. OCH Hip А #ра Find the best wait for this hypothes round to two decimal places as needed) Duthorthal Round love decimal places needed dance of OA Do not O Do not There ist There is not conce once c Use the given statistics to complete parts (a) and (b). Assume that the populations are normally distributed (a) Test whether at the 0.10 level of significance for the given sample data. (1) Construct a 90% confidence interval about 2 H2 на на Find the test statistic for this hypothesis test. A (Round to two decimal places as needed) Determine the P-value for this hypothesis test (Round to three decimal places as needed) State the conclusion for this hypothesis tost O A Do not reject There is sufficient evidence at the 10 level of significance to conclude that OB. Do not reject There is not sufficient evidence at the 0.10 level of significance to conclude that O O C. Reject Ho There is not sufficient evidence at the 0.10 level of significance to conclude that D. Raject Ho There is sufficient evidence at the 0.10 level of significance to conclude that is the range from a lower bound of to an upper bound of (b) The 90% confidence interval about (Round to three decimal places as needed) Click to select your answer(s)

Question William swims a 100 meter race in 90.7 seconds, on average, with a standard deviation of 5. He decided to see if using a new breathing technique improves his time. He used the new technique for 10 races. For those 10 races, William got an average of 87.4 seconds. He wants to know if the new technique helped him swim a faster race. Which of the answer choices shows the correct first three steps of a one-mean hypothesis test at the 1% significance level? Select the correct answer below: O H := 90.7; H: > 90.7, which is a right-tailed test. a=0.01. 20 = 874-90.7 -2.09. VO O H : 4 = 87.4; H :N < 87.4, which is a left-tailed test. a = 0.01. 90.7–87. 4 2.09. 5 O H := 90.7; H : 4 < 90.7, which is a left-tailed test. a=0.01. 90.7-87.4 0.74. 20 O H :u= 87.4; H:> 87.4, which is a right-tailed test. a=0.01. 90.7–87.4 2.09. O H : = 90.7; H.: > 90.7, which is a right-tailed test. a=0.01. 20 = 0 87.4–90.7 -0.74. O H :u= 90.7; H :N < 90.7, which is a left-tailed test. a=0.01. 20 = 87.4–90.7 -2.09.

Ariana gets a distance of 15.6 meters, on average, for her shot put throws, with a standard deviation of 1. She decided to see if using a new throwing technique would increase her distance. She used the new technique for 12 throws. For those 12 throws, Ariana got an average of 16.1 meters. She wants to know if the new technique helped her get more distance on each throw. Which of the answer choices shows the correct first three steps of a one-mean hypothesis test at the 5% significance level? Select the correct answer below: O H u < 16.1, which is a left-tailed test. : u = 16.1; H a = 0.05. - 15.6-16.1 -1.73. H :u= 15.6; H:> 15.6, which is a right-tailed test. a = 0.05. 2 = 16.1-15.6 1.73. H:u= 15.6; H :N < 15.6, which is a left-tailed test. a = 0.05. 2 = 15.6-16.1 -0.04. O H , :u = 16.1; H, : > 16.1, which is a right-tailed test. a = 0.05. = 15,6-16.1 -1.73. H :u= 15.6; H:<15.6, which is a left-tailed test. a=0.05. 15.6-16.1 -0.04. 20 = 20 15.6-16,1 H :u= 16.1; H:4 > 16.1, which is a right-tailed test. a = 0.05. - 15.6-16.1 -1.73. H: = 15.6; H :N > 15.6, which is a right-tailed test. a = 0.05. = 16.1-15. 6 00 H :u= 15.6; H: 4 < 15.6, which is a left-tailed test. a = 0.05. A = 16.1-15.6 1.73. VII

Ariana continues her hypothesis test, by finding the p-value to make a conclusion about the null hypothesis. Ho : µ = 15.6; Ha : µ > 15.6, which is a right-tailed test. a = 0.05. • zo = 1.73 Which is the correct conclusion of Ariana's one-mean hypothesis test at the 5% significance level? 0.05 0.07 0.00 0.01 0.02 0.03 0.04 0.06 0.08 0.09 0.9484 0.9495 0.9452 0.9463 0.9474 0.9505 0.9515 0.9525 0.9535 0.9545 1.6 0.9591 0.9625 0.9564 0.9633 1.7 0.9554 0.9573 0.9582 0.9599 0.9608 0.9616 0.9699 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9706 1.8 0.9738 0.9713 0.9719 0.9726 0.9732 0.9744 0.9750 0.9756 0.9761 0.9767 1.9 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 0.9817 2.0 Use the above portion of the Standard Normal Table. Select the correct answer below: p= 0.0418 We should not reject Ho because p 2 a. So, at the 5% significance level, the data do not provide sufficient evidence to conclude that the new technique helped Ariana get more distance on each throw. p = 0.0418 We should reject Ho because p < a. So, at the 5% significance level, the data provide sufficient evidence to conclude that the new technique helped Ariana get more distance on each throw. p = 0.9582 We should not reject Ho because p> a. So, at the 5% significance level, the data do not provide sufficient evidence to conclude that the new technique helped Ariana get more distance on each throw. p = 0.9582 We should reject Ho because p < a. So, at the 5% significance level, the data provide sufficient evidence to conclude that the new technique helped Ariana get more distance on each throw.

(a) Use a = 0.10 to test for a statistically significantly difference between the population means for first- and fourth-round scores. State the null and alternative hypotheses. (Use ua = mean score first round - mean score fourth round.) 0 0 Houd > 0 Haug so Ho: un so Hailg>0 Ho: шағО Haug = 0 O Ho: Mo = 0 Hailloso • Ho: Mo = 0 Hauro Calculate the value of the test statistic. (Round your answer to three decimal places.) Calculate the p-value. (Round your answer to four decimal places.) p-value = State your conclusion. • Do not reject H. There is insufficient evidence to conclude that the mean score for the first round of a golf tournament is significantly different than the mean score for the fourth and final round. Reject Ho. There is insufficient evidence to conclude that the mean score for the first round of a golf tournament is significantly different than the mean score for the fourth and final round. O Do not Reject Ho. There is sufficient evidence to conclude that the mean score for the first round of a golf tournament is significantly different than the mean score for the fourth and final round. Reject Ho. There is sufficient evidence to conclude that the mean score for the first round of a golf tournament is significantly different than the mean score for the fourth and final round. (b) What is the point estimate of the difference between the two population means? (Use mean score first round - mean score fourth round.) For which round is the population mean score lower? The mean of the fourth round scores was lower than the mean of the first round scores. The mean of the first round scores was lower than the mean of the fourth round scores.

Assume that the differences are normally distributed. Complete parts (a) through (d) below. Observation 1 53.0 54.6 2 51.4 51.7 3 42.7 47.2 4 42.9 47.6 5 51.6 522 6 48.2 49.8 7 45. 9 46.6 8 48.6 49.7 3 4 5 6 (a) Determine di = X; - Y, for each pair of data. Observation 1 2 -1.6 -0.3 (Type integers or decimals.) (b) Computed and so 7 -0.7 8 - 1.1 di -4.5 -45 -47 -4.7 -0.6 -1.6 a = - 1.888 (Round to three decimal places as needed.) $a = (Round to three decimal places as needed.) (c) Test if He <0 at the a = 0.05 level of significance. What are the correct null and alternative hypotheses? O A. Ho: Ho<0 H:Hd = 0 | 0C. Ho Ha = 0 H:Hd <0 OB. Ho Hd < 0 H:Hd > 0 OD. Ho: Ma>0 | H: Ha < 0 P-value = (Round to three decimal places as needed.) Choose the correct conclusion below. O A. Do not reject the null hypothesis. There is insufficient evidence that Hd <0 at the a= 0.05 level of significance. OB. Reject the null hypothesis. There is sufficient evidence that Hd <0 at the a=0.05 level of significance. O C. Do not reject the null hypothesis. There is sufficient evidence that a <0 at the a = 0.05 level of significance. OD. Reject the null hypothesis. There is insufficient evidence that Hd <0 at the a=0.05 level of significance (d) Compute a 95% confidence interval about the population mean difference Hd The lower bound is The upper bound is (Round to two decimal places as needed.)

> A Moving to another question will save this response. Question 49 The test statistic for a certain left sided hypothesis test is zo=.40. At the 10% significance level, which of the following is false? A. There is sufficient evidence to reject the null hypothesis B. P-value > significance level C. The test is not statistically significant D. The correct answer is not among the choices. E. The sample mean exceeds the mean stated in the null hypothesis > Moving to another question will save this response. 4000927 201