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rahulv336699Vellore Institute of Technology - VIT

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Answer: Step-by-step explanation: Average is the sum of all observations divid...

 

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.

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Answer: Step-by-step explanation: Given that,Point estimate = sample mean = = ...
Answer: Step-by-step explanation: first o all calculating the mean and standar...
Answer: Step-by-step explanation: The margin of error (ME) is . a) When , b) W...
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
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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)
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Answer: Step-by-step explanation: Given that, = 101.0 = 99.5 = 3.1 = 5.0 n1 = ...
Answer: Step-by-step explanation: Given that , mean = = 50 standard deviation ...
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Answer: Sample size = 11
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Answer: Step-by-step explanation: Step 1 of 3 From the information, observe th...
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%
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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%
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Answer: Step-by-step explanation: A college professor states that this year's ...
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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
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Answer: Step-by-step explanation:Ho: p = 0.68H1: p is not equal to 0.68
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
Answer: Step-by-step explanation: Test Results
Answer: Step-by-step explanation: Note: There is no need of degrees of freedom...
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Answer: Step-by-step explanation: Test statistic = -0.309
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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.
Answer: Step-by-step explanation: According to the given probability table of ...
(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.
Answer: Step-by-step explanation: (a) H0: = 0 Ha: 0 Values of d are given by: ...
Answer: Step-by-step explanation: from above: sd =1.737 p value =0.009 d) for ...
Answer: Step-by-step explanation: Here the sample mean exceeds the population ...

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