Which of the following are true statements about hypothesis tests to compare two

Question

Which of the following are true statements about hypothesis tests to compare two population means? The Z-distribution is used to obtain the critical value for the rejection region and decision rule whenever at least one of the sample size is sufficiently large (either n_1 or n_2 is at least 30). D_0 is the amount by which the population means are hypothesized to differ. Although this amount if often 0(if the means are hypothesized to be equal),it does not have to be zero. Both A and B. Neither A or B

Explanation / Answer

B is the answer.

Formulate an Analysis Plan

The analysis plan describes how to use sample data to accept or reject the null hypothesis. It should specify the following elements.

Analyze Sample Data

Using sample data, find the standard error, degrees of freedom, test statistic, and the P-value associated with the test statistic.

SE = sqrt[ (s12/n1) + (s22/n2) ]

DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] }

t = [ (x1 - x2) - d ] / SE

Interpret Results

If the sample findings are unlikely, given the null hypothesis, the researcher rejects the null hypothesis. Typically, this involves comparing the P-value to the significance level, and rejecting the null hypothesis when the P-value is less than the significance level.

Test Your Understanding

In this section, two sample problems illustrate how to conduct a hypothesis test of a difference between mean scores. The first problem involves a two-tailed test; the second problem, a one-tailed test.

Problem 1: Two-Tailed Test

Within a school district, students were randomly assigned to one of two Math teachers - Mrs. Smith and Mrs. Jones. After the assignment, Mrs. Smith had 30 students, and Mrs. Jones had 25 students.

At the end of the year, each class took the same standardized test. Mrs. Smith's students had an average test score of 78, with a standard deviation of 10; and Mrs. Jones' students had an average test score of 85, with a standard deviation of 15.

Test the hypothesis that Mrs. Smith and Mrs. Jones are equally effective teachers. Use a 0.10 level of significance. (Assume that student performance is approximately normal.)

Solution: The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: 1 - 2 = 0
Alternative hypothesis: 1 - 2 0

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.10. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t-score test statistic (t).

SE = sqrt[(s12/n1) + (s22/n2)]
SE = sqrt[(102/30) + (152/25] = sqrt(3.33 + 9) = sqrt(12.33) = 3.51

DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] }
DF = (102/30 + 152/25)2 / { [ (102 / 30)2 / (29) ] + [ (152 / 25)2 / (24) ] }
DF = (3.33 + 9)2 / { [ (3.33)2 / (29) ] + [ (9)2 / (24) ] } = 152.03 / (0.382 + 3.375) = 152.03/3.757 = 40.47

t = [ (x1 - x2) - d ] / SE = [ (78 - 85) - 0 ] / 3.51 = -7/3.51 = -1.99

where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.

Since we have a two-tailed test, the P-value is the probability that a t-score having 40 degrees of freedom is more extreme than -1.99; that is, less than -1.99 or greater than 1.99.

We use the t Distribution Calculator to find P(t < -1.99) = 0.027, and P(t > 1.99) = 0.027. Thus, the P-value = 0.027 + 0.027 = 0.054.

Interpret results. Since the P-value (0.054) is less than the significance level (0.10), we cannot accept the null hypothesis.

Note: If you use this approach on an exam, you may also want to mention why this approach is appropriate. Specifically, the approach is appropriate because the sampling method was simple random sampling, the samples were independent, the sample size was much smaller than the population size, and the samples were drawn from a normal population.

Problem 2: One-Tailed Test

The Acme Company has developed a new battery. The engineer in charge claims that the new battery will operate continuously for at least 7 minutes longer than the old battery.

To test the claim, the company selects a simple random sample of 100 new batteries and 100 old batteries. The old batteries run continuously for 190 minutes with a standard deviation of 20 minutes; the new batteries, 200 minutes with a standard deviation of 40 minutes.

Test the engineer's claim that the new batteries run at least 7 minutes longer than the old. Use a 0.05 level of significance. (Assume that there are no outliers in either sample.)

Solution: The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: 1 - 2 >= 7
Alternative hypothesis: 1 - 2 < 7

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the mean difference between sample means is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t-score test statistic (t).

SE = sqrt[(s12/n1) + (s22/n2)]
SE = sqrt[(402/100) + (202/100] = sqrt(16 + 4) = 4.472

DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] }
DF = (402/100 + 202/100)2 / { [ (402 / 100)2 / (99) ] + [ (202 / 100)2 / (99) ] }
DF = (20)2 / { [ (16)2 / (99) ] + [ (2)2 / (99) ] } = 400 / (2.586 + 0.162) = 145.56

t = [ (x1 - x2) - d ] / SE = [(200 - 190) - 7] / 4.472 = 3/4.472 = 0.67

where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between population means, and SE is the standard error.

Here is the logic of the analysis: Given the alternative hypothesis (1 - 2 < 7), we want to know whether the observed difference in sample means is small enough (i.e., sufficiently less than 7) to cause us to reject the null hypothesis.

The observed difference in sample means (10) produced a t-score of 0.67. We use the t Distribution Calculator to find P(t < 0.67) = 0.75.

This means we would expect to find an observed difference in sample means of 10 or less in 75% of our samples, if the true difference were actually 7. Therefore, the P-value in this analysis is 0.75.

Interpret results. Since the P-value (0.75) is greater than the significance level (0.05), we cannot reject the null hypothesis.

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