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Chapter 11: Problem 13

(a) state the decision rule, (b) compute the pooled estimate of the population variance, (c) compute the test statistic, (d) state your decision about the null hypothesis, and (e) estimate the \(p\) -value. The null and alternate hypotheses are: $$ \begin{array}{l} H_{0}: \mu_{1}=\mu_{2} \\ H_{1}: \mu_{1} \neq \mu_{2} \end{array} $$ A random sample of 10 observations from one population revealed a sample mean of 23 and a sample deviation of \(4 .\) A random sample of 8 observations from another population revealed a sample mean of 26 and a sample standard deviation of \(5 .\) At the .05 significance level, is there a difference between the population means?

Short Answer

Expert verified
Do not reject the null hypothesis; no significant difference between means.

Step by step solution

01

State the Decision Rule

To determine the decision rule, we need to know the critical value at the 0.05 significance level for a two-tailed test. Since the sample sizes are small, we'll use the t-distribution. The degrees of freedom for the t-distribution can be calculated using the formula:\[ n_1 + n_2 - 2 = 10 + 8 - 2 = 16 \]Using the t-distribution table, the critical t-value (two-tailed) for 16 degrees of freedom at the 0.05 significance level is approximately 2.120. We will reject the null hypothesis if the absolute value of the calculated test statistic is greater than 2.120.
02

Compute the Pooled Estimate of the Population Variance

The pooled estimate of the population variance \( S_p^2 \) is calculated using the following formula:\[ S_p^2 = \frac{(n_1 - 1)S_1^2 + (n_2 - 1)S_2^2}{n_1 + n_2 - 2} \]Substitute the given values:\[ S_p^2 = \frac{(10 - 1)(4^2) + (8 - 1)(5^2)}{10 + 8 - 2} \]\[ S_p^2 = \frac{9(16) + 7(25)}{16} \]\[ S_p^2 = \frac{144 + 175}{16} \]\[ S_p^2 = \frac{319}{16} = 19.9375 \]
03

Compute the Test Statistic

The test statistic \( t \) for comparing two means is calculated using the formula:\[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{S_p^2 \left( \frac{1}{n_1} + \frac{1}{n_2} \right) }} \]Substitute the given values and the pooled variance:\[ t = \frac{23 - 26}{\sqrt{19.9375 \left( \frac{1}{10} + \frac{1}{8} \right) }} \]\[ t = \frac{-3}{\sqrt{19.9375 \left( 0.125 \right) }} \]\[ t = \frac{-3}{\sqrt{2.4922}} \]\[ t = \frac{-3}{1.578} \approx -1.901 \]
04

State Your Decision About the Null Hypothesis

Compare the test statistic with the critical value from Step 1. Since the absolute value of the test statistic (1.901) is less than the critical t-value (2.120), we do not reject the null hypothesis. There is not enough evidence to suggest a difference between the population means at the 0.05 level of significance.
05

Estimate the p-Value

Using a t-distribution table or software, we find the p-value associated with the test statistic of -1.901 and 16 degrees of freedom. The p-value is found to be approximately 0.075. Since the p-value is greater than the significance level of 0.05, we fail to reject the null hypothesis.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Decision Rule
In hypothesis testing, establishing a decision rule is a crucial step. It determines the criteria for rejecting or failing to reject the null hypothesis. For a two-tailed test at the 0.05 significance level, the critical value from the t-distribution is used. This is because our sample sizes are small, requiring us to rely on the t-distribution rather than the normal distribution.

For 16 degrees of freedom (calculated as the sum of sample sizes minus 2), we find that the critical t-value is approximately 2.120. This value serves as our benchmark. If the absolute test statistic exceeds 2.120, we reject the null hypothesis. Otherwise, we do not have sufficient evidence to do so. Deciding whether or not to reject the null hypothesis depends on the comparison between this critical value and our calculated test statistic.
Test Statistic
The test statistic provides a standardized way to determine how far the sample means are from each other under the null hypothesis. For two independent samples, the test statistic formula helps to highlight the difference between two sample means and considers variability within the samples.

The formula used is: \[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{S_p^2 \left( \frac{1}{n_1} + \frac{1}{n_2} \right) }} \]where \( \bar{x}_1 \) and \( \bar{x}_2 \) are the sample means, and \( S_p^2 \) is the pooled variance.

In the provided example, we substitute the given values to get a test statistic of approximately -1.901. This reflects how many standard errors the difference in sample means is away from zero, helping us decide if this difference is statistically significant.
P-value Estimation
The p-value quantifies the probability of observing a test statistic at least as extreme as the one calculated, given that the null hypothesis is true. It is used to determine the statistical significance of the test.

A smaller p-value indicates stronger evidence against the null hypothesis. For the given test statistic and degrees of freedom, we use statistical tables or software to find an approximate p-value.

In our exercise, with a test statistic of -1.901 and 16 degrees of freedom, the p-value is about 0.075. Because this is greater than our significance level of 0.05, we fail to reject the null hypothesis. This suggests there is not enough evidence to state there is a significant difference between the population means.
Pooled Population Variance
The pooled population variance provides a weighted estimate of variance assuming equal population variances across the two samples. It leverages both sample variabilities for a more accurate estimate.

We use the formula: \[ S_p^2 = \frac{(n_1 - 1)S_1^2 + (n_2 - 1)S_2^2}{n_1 + n_2 - 2} \]to combine the variances of both samples.

In our example, after substituting the respective sample sizes and variances (which are squared), the pooled variance is calculated as 19.9375. This value is then used in further calculations, including the test statistic determination, to properly weigh and compare the two sample means under the null hypothesis.

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Most popular questions from this chapter

Mary Jo Fitzpatrick is the vice president for Nursing Services at St. Luke's Memorial Hospital. Recently she noticed in the job postings for nurses that those that are unionized seem to offer higher wages. She decided to investigate and gathered the following information. $$ \begin{array}{|lccc|} \hline & & \text { Population } & \\ \text { Group } & \text { Mean Wage } & \text { Standard Deviation } & \text { Sample Size } \\ \hline \text { Union } & \$ 20.75 & \$ 2.25 & 40 \\ \text { Nonunion } & \$ 19.80 & \$ 1.90 & 45 \\ \hline \end{array} $$ Would it be reasonable for her to conclude that union nurses earn more? Use the .02 significance level. What is the \(p\) -value?

Clark Heter is an industrial engineer at Lyons Products. He would like to determine whether there are more units produced on the night shift than on the day shift. Assume the population standard deviation for the number of units produced on the day shift is 21 and is 28 on the night shift. A sample of 54 day-shift workers showed that the mean number of units produced was 345\. A sample of 60 night-shift workers showed that the mean number of units produced was \(351 .\) At the .05 significance level, is the number of units produced on the night shift larger?

A sample of 40 observations is selected from one population with a population standard deviation of \(5 .\) The sample mean is \(102 .\) A sample of 50 observations is selected from a second population with a population standard deviation of \(6 .\) The sample mean is \(99 .\) Conduct the following test of hypothesis using the .04 significance level. $$ \begin{array}{l} H_{0}: \mu_{1}=\mu_{2} \\ H_{1}: \mu_{1} \neq \mu_{2} \end{array} $$ a. Is this a one-tailed or a two-tailed test? b. State the decision rule. c. Compute the value of the test statistic. d. What is your decision regarding \(H_{0}\) e. What is the \(p\) -value?

The Roper Organization conducted identical surveys 5 years apart. One question asked of women was "Are most men basically kind, gentle, and thoughtful?" The earlier survey revealed that, of the 3,000 women surveyed, 2,010 said that they were. The later revealed 1,530 of the 3,000 women surveyed thought that men were kind, gentle, and thoughtful. At the .05 level, can we conclude that women think men are less kind, gentle, and thoughtful in the later survey compared with the earlier one?

A coffee manufacturer is interested in whether the mean daily consumption of regular-coffee drinkers is less than that of decaffeinated-coffee drinkers. Assume the population standard deviation for those drinking regular coffee is 1.20 cups per day and 1.36 cups per day for those drinking decaffeinated coffee. A random sample of 50 regular-coffee drinkers showed a mean of 4.35 cups per day. A sample of 40 decaffeinated-coffee drinkers showed a mean of 5.84 cups per day. Use the .01 significance level. Compute the \(p\) -value.
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