🏷️ LIMITED TIME OFFER: GET 20% OFF GRADE+ YEARLY SUBSCRIPTION →
Log in
HomeStudy Guides420,000CA160,000

CHYS 3P15 : CHYS 3P15 - Lecture Notes.docx

192 views15 pages
carminecamel732
19 Apr 2012
School
Brock University
Department
Child and Youth Studies
Course
CHYS 3P15
Professor
Heather Lawford

For unlimited access to Study Guides, a Grade+ subscription is required.

Document Summary

Steps in hypothesis testing: develop null & alternative hypothesis, set alpha, calculate power to determine sample size, collect data, collect statistic and p-value, accept or reject the null. Sampling distribution is a frequency distribution of a sample statistic. Distribution of sample means is usually referred to as the sampling distribution of means, or the sampling distribution of the mean. All hypotheses test are comprised of 2 mutually exclusive hypothesis. Null: both means come from the same population: any difference between the means is due to chance. Null: control = experimental: this is the hypothesis that you are able to test, if the null hypothesis is probably not true, what must we conclude. Control experimental: directional hypothesis (1 tailed) Experimental > control cant test this hypothesis directly, but can find support by rejecting the null hypothesis. We cannot just look at means and determine if they differ remember random error statistical fluctuations in the measured data random error is always present.

Get access

Grade+20% off
$8 USD/m$10 USD/m
Billed $96 USD annually
Grade+
Homework Help
Study Guides
Textbook Solutions
Class Notes
Textbook Notes
Booster Class
40 Verified Answers

Related Documents

CHYS 3P15 Lecture Notes - Lecture 7: Null Hypothesis, Type I And Type Ii Errors, Alternative Hypothesis
chocolatemouse764
CHYS 3P15
Final Exam
Study Guide
CHYS 3P15 Final: CHYS 3P15 - Textbook Review for Exam.docx
carminecamel732
CHYS 3P15 Lecture Notes - Lecture 4: Null Hypothesis
violetoyster687

Related Questions

1 Which of the following statements about Type I and Type II errors is correct
a Type I: Reject a true alternative hypothesis. Type II: Do not reject a false alternative.
b Type I: Reject a true null hypothesis. Type II: Do not reject a false null hypothesis.
c Type I: Reject a false null hypothesis. Type II: Reject a true null hypothesis.
d Type I: Do not reject a false null hypothesis. Type II: Reject a true null hypothesis.
2 You are reading a report that contains a hypothesis test you are interested in. The writer of the report writes that the p-value for the test you are interested in is 0.0749, but does not tell you the value of the test statistic. From this information you can:
a Not reject the hypothesis at a Probability of Type I error = .05, but reject the hypothesis at a Probability of Type I error = 0.10
b Reject the hypothesis at a Probability of Type I error = .05, and reject at a Probability of Type I error = 0.10
c Not reject the hypothesis at a Probability of Type I error = 0.05, and not reject at a Probability of Type I error = 0.10
d Reject the hypothesis at a Probability of Type I error = .05, but not reject at a Probability of Type I error = 0.10
3 The random sample below is obtained to test the following hypothesis about the population mean.
H?: ? ? 1500
H?: ? > 1500
620 1711 366 2528 2678 1661 442 725 1938
409 330 2480 542 369 2124 549 2074 1665
1873 873 2143 2061 1177 2509 1264 2397 1523
1837 1958 1041 1639 2199 2232 387 2270 2136
1111 1883 2612 2230 1597 1726 694 1990 1354
2090 909 2128 1608 747 1121 2220 2390 2347
1041 316 655 632 2064 1901 532 552 846
2704 1410 2165 1065 937 1452 2539 410 656
1169 527 809 2364 2350 2210 1459 2391 856
2711 1985 2382 2289 1927 518 2177 437 1151
2018 1580 607 2715 2188 1691 1394 2610 1186
695 2428 2246 858 2036 1681 2449 1578 1971
1846 1729 2389 1737 1913 1863 2072 2593 2287
2220 2230 551 458 2626 2731 488 2551 1736
1373 307 1803 2647 2679 1508 1468 1443 516
1002 2116 2616 817 2522 460 1879 1999 1837
The level of significance of the test is ? = 0.05. Compute the relevant test statistic.
This is a(n) _______ (two-tail, upper-tail, lower-tail) test. The test statistic is TS = _______.
a Two-tail test TS = 1.81
Do not reject H?: ? ? 1500. Conclude that the population mean is not greater than 1500.
b Upper tail test. TS = 1.52
Do not reject H?: ? ? 1500. Conclude that the population mean is not greater than 1500.
c Upper tail test. TS = 1.81
Reject H?: ? ? 1500. Conclude that the population mean is greater than 1500.
d Lower tail test. TS = 1.98
Do not reject H?: ? ? 1500. Conclude that the population mean is no greater than 1500.
4 Consider the following hypothesis test.
H?: ? ? 30
H?: ? > 30
A random sample of n = 15 yielded the following observations
51 38 26 16 28
57 20 33 35 23
21 47 56 54 36
Use ? = 0.05
TS = ______ CV = ______ State the decision rule.
a 1.68 1.761 Do not reject H?. Conclude the mean is not greater than 30.
b 1.68 1.64 Reject H?. Conclude the mean is greater than 30.
c 1.847 2.145 Do not reject H?. Conclude the mean is not less than 30.
d 1.847 1.761 Reject H?. Conclude the mean is less than 30.
5 In a recent study, a major fast food restaurant had a mean service time of 165 seconds. The company embarks on a quality improvement effort to reduce the service time and has developed improvements to the service process. The new process will be tested in a sample of stores. The new process will be adopted in all of its stores, if it reduced mean service time by more than 45 seconds compared to the current mean service time. To perform the hypothesis test, the sample of 48 stores yields the following data (seconds).
90 96 133 108 136 110 119 138
129 98 101 92 135 124 115 90
132 125 110 124 126 138 94 130
108 96 140 135 102 114 109 137
138 104 108 134 92 107 96 119
105 111 96 136 126 116 98 131
Use ? = 0.05
|TS| = ______ |CV| = ______
a 1.548 1.678 Do not reject H?. The mean service time is not reduced by more than 45 seconds. Do not adopt the new process.
b 1.871 1.678 Reject H?. The mean service time is reduced by more than 45 seconds. Adopt the new process.
c 1.871 1.640 Do not reject H?. The mean service time is not reduced by more than 45 seconds. Do not adopt the new process.
d 1.548 1.640 Reject H?. The mean service time is reduced by more than 45 seconds. Adopt the new process.

Introduction: A Chi-square test is used to compare observed data with expected data according to a hypothesis. For instance, if you were crossbreeding 2 heterozygous pea plants, you would expect to see a 3:1 phenotypic ratio in the offspring. In this case, if you were to breed 400 pea plants, you would expect to see 300 plants showing the dominant trait and 100 showing the recessive trait. But what happens if you observe only 260 plants with the dominant trait and 140 plants with the recessive trait? Does this mean something is wrong with Mendelian genetics or is this difference in expected results just due to chance (random sampling error)? These are the questions that can be answered using Chi-square statistics. The results of this statistical test is used to either reject or accept (fail to reject) the null hypothesis. The null hypothesis states there is no significant difference between the observed results and the expected results. This means that if the null hypothesis is accepted, the difference in observed and expected results was just a matter of chance and so the observed results basically "fit" with what was expected. Degrees of freedom (df) = number of independent outcomes (Y) being compared less 1 df = Y-1 At the 95% confidence interval we are 95% confident that there is a significant difference between the observed and expected results, therefore rejecting the null hypothesis. Probability Value - Is the decimal value determined from the X2 table and is the probability of accepting the null hypothesis. A 0.05 probability value equates to a 95% confidence interval.

The Chi-squared test formula is: Example: If we cross two pea plants that are heterozygous yellow pods, we would expect a 3:1 phenotypic ratio. So let's say we actually did the cross and got 280 plants with green pods and 120 plants with yellow pods. Question: Is this a 3:1 phenotypic ratio? This is the value of Chi-squared Test. We have a total of 400 plants and we expect a 300 green:100 yellow phenotypic ratio If the calculated Chi-squared value is less than the critical value listed in the Chi-squared table, then we accept the null hypothesis. This means that there is no significant difference between the observed and the expected values. Our degrees of freedom (df) = 2 outcomes - 1, or df = 1. Now we go the X2 table below and using the df = 1 and probability value of 0.05, our critical value is 3.84. Since our calculated X2 value is 5.33, and is larger than the critical value, we reject the null hypothesis and can say (at 95% confidence) that there is a significant difference between our observed and expected values.

The parent generation is yellowed podded and green podded pea plants. You cross a yellow podded pea plant with a green podded pea plant and you get 100% yellow podded plants in the F1 Generation (Phenotypic ratio 4 : 0, yellow to green). What will be the expected phenotypic ratio when you allow the F1 generation to reproduce?

Fill out the Punnett square.

If we actually did the cross and got 1150 yellow and 350 green. Would this be a consistent with what was expected?

Learning Outcomes Questions

1. Why would you run a Chi-squared test?

To determine if our data is consistent with expected results.

a To determine if our data is consistent with expected results.

b To determine if our data exactly matches the expected results.

c To determine the expected results.
d

To compare the phenotypic ratios to the genotypic ratios.

2. Determine the degrees of Freedom of the phenotypic ratio for this genetic cross.

a. 1

b. 2

c. 3

d. 4

e. 5

3. Using the data given, what is the result of your Chi-squared analysis? x2= ___.

a.

2.22
b

2.71
c

4.36
d

187.78
e

448.27

4. Using the results of your Chi-squared analysis, do we fail to reject or reject the null hypothesis?

a.

Fail to reject the null
b.

Reject the null
c.

It cannot be determined from the data given

Hi - could someone please assist me with this question? It's giving me nightmares......thanks

Historically, the average waiting time spent in a queue on Friday afternoon at a bank's city branch was 10 minutes. To determine whether their new multi-queuing system constitutes an improvement, a random sample of 15 customers was taken and their waiting time recorded. The results are in the X1 column of the datafile P14.12. Assume the distribution of waiting times is normally distributed.

1. State the direction of the alternative hypothesis used to test whether there is an improvement in average waiting time. Type gt (greater than), ge (greater than or equal to), lt (less than), le (less than or equal to) or ne (not equal to) as required in the box.


2. Calculate the test statistic correct to three decimal places (hint: use Descriptive Statistics to calculated the standard deviation and sample mean).


3. Use Kaddstat to determine the p-value for the test correct to four decimal places.


4. Is the null hypothesis rejected for this test if a 10% level of significance is used? Type yes or no.


5. Disregarding your answer for 4, if the null hypothesis was rejected, could we conclude that the average waiting time has improved with the new queuing system? Type yes or no.

DATA FILE

135