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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
Fail to reject the null hypothesis; no evidence of a difference in means.

Step by step solution

01

Identify Hypotheses

The null hypothesis \(H_0\) is that the population means are equal: \(\mu_1 = \mu_2\). The alternate hypothesis \(H_1\) suggests that the means are not equal: \(\mu_1 eq \mu_2\).
02

Calculate Test Statistic

Use the formula for the standard error of the difference between two means: \(SE = \sqrt{ \frac{S_1^2}{n_1} + \frac{S_2^2}{n_2} }\), where \(S_1 = 4, \ n_1 = 10, \ S_2 = 5, \ n_2 = 8\). This gives \(SE = \sqrt{ \frac{4^2}{10} + \frac{5^2}{8} } = \sqrt{ 1.6 + 3.125 } = \sqrt{4.725} \approx 2.173\). Then calculate the test statistic: \( t = \frac{\bar{x}_1 - \bar{x}_2}{SE} = \frac{23 - 26}{2.173} = \frac{-3}{2.173} \approx -1.38\).
03

Determine Critical Value

Since this is a two-tailed test at the 0.05 significance level with degrees of freedom approximated by \(df = \min(n_1 - 1, n_2 - 1) = \min(9, 7) = 7\), use a t-distribution table to find the critical t-value for \(df = 7\). The critical value is approximately \(\pm 2.3646\).
04

Make Decision

Compare the absolute value of the test statistic \(|-1.38|\) to the critical value \(2.3646\). Since \(1.38 < 2.3646\), we fail to reject the null hypothesis. There is not enough evidence to conclude that there is a significant difference between the population means at the 0.05 level.

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Key Concepts

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

Null and Alternate Hypotheses
In hypothesis testing, we need to clearly define two hypotheses: the null hypothesis and the alternate hypothesis. These form the foundation of our test.

The **null hypothesis** (\( H_0 \)) proposes that there is no effect or difference, and in this exercise, it asserts that the population means from two different samples are equal. It is expressed as \( \mu_1 = \mu_2 \), suggesting that any observed difference is due to random chance.

Conversely, the **alternate hypothesis** (\( H_1 \)) suggests that there is indeed an effect or a difference between the two population means. It counterclaims the null hypothesis by stating \( \mu_1 eq \mu_2 \), indicating that the means are not equal.
  • **Null Hypothesis (\( H_0 \)):** No difference.\( \mu_1 = \mu_2 \).
  • **Alternate Hypothesis (\( H_1 \)):** There is a difference.\( \mu_1 eq \mu_2 \).
Understanding these hypotheses is crucial as they guide the selection of the statistical test and influence the interpretation of the results.
Test Statistic Calculation
Calculating the test statistic is a key step in hypothesis testing, acting like the evidence you provide to support or refute your hypotheses. The test statistic tells you how many standard deviations your sample mean is away from the null hypothesis mean.

**Step-by-Step Test Statistic Calculation:**
  • Calculate the **standard error** of the difference between the means.\[ SE = \sqrt{ \frac{S_1^2}{n_1} + \frac{S_2^2}{n_2} } \] With given values: \( S_1 = 4 \), \( n_1 = 10 \), \( S_2 = 5 \), \( n_2 = 8 \),\[ SE = \sqrt{ \frac{4^2}{10} + \frac{5^2}{8} } = \sqrt{ 4.725 } \approx 2.173 \]
  • Determine the **test statistic**:\[ t = \frac{\bar{x}_1 - \bar{x}_2}{SE} = \frac{23 - 26}{2.173} \approx -1.38 \]
This calculation combines sample means, standard deviations, and sample sizes to articulate how unlikely the observed sample difference is, assuming that the null hypothesis is true.
Significance Level
The significance level is a criterion used to decide whether to reject the null hypothesis. Commonly denoted by \( \alpha \), it quantifies the risk of incorrectly rejecting \( H_0 \). Typical significance levels are 0.05, 0.01, and 0.10.

In this exercise, we use a 0.05 significance level, which means there's a 5% risk of concluding there is a difference when there is none. It also defines the rejection region, dividing the probability distribution into a critical region: if the test statistic falls into this region, we reject \( H_0 \)
  • At \( \alpha = 0.05 \), the critical t-value (two-tailed) provides boundaries for the decision rule.
  • For degrees of freedom \( df = 7 \), the critical value is approximately \( \pm 2.3646 \).
With the calculated test statistic \( t = -1.38 \), we compare it against \( \pm 2.3646 \). Without reaching the critical value, we do not reject \( \mu_1 = \mu_2 \), implying insufficient evidence to declare a substantial difference.

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Most popular questions from this chapter

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 15 observations from the first population revealed a sample mean of 350 and a sample standard deviation of \(12 .\) A random sample of 17 observations from the second population revealed a sample mean of 342 and a sample standard deviation of \(15 .\) At the .10 significance level, is there a difference in the population means?

A computer manufacturer offers a help line that purchasers can call for help 24 hours a day 7 days a week. Clearing these calls for help in a timely fashion is important to the company's image. After telling the caller that resolution of the problem is important the caller is asked whether the issue is "software" or "hardware" related. The mean time it takes a technician to resolve a software issue is 18 minutes with a standard deviation of 4.2 minutes. This information was obtained from a sample of 35 monitored calls. For a study of 45 hardware issues, the mean time for the technician to resolve the problem was 15.5 minutes with a standard deviation of 3.9 minutes. This information was also obtained from monitored calls. At the . 05 significance level is it reasonable to conclude that it takes longer to resolve software issues? What is the \(p\) -value?

The null and alternate hypotheses are: $$\begin{array}{l}H_{0}: \pi_{1}=\pi_{2} \\\H_{1}: \pi_{1} \neq \pi_{2}\end{array}$$ A sample of 200 observations from the first population indicated that \(X_{1}\) is \(170 .\) A sample of 150 observations from the second population revealed \(X_{2}\) to be \(110 .\) Use the .05 significance level to test the hypothesis. a. State the decision rule. b. Compute the pooled proportion. c. Compute the value of the test statistic. d. What is your decision regarding the null hypothesis?

Each month the National Association of Purchasing Managers publishes the NAPM index. One of the questions asked on the survey to purchasing agents is: Do you think the economy is expanding? Last month, of the 300 responses 160 answered yes to the question. This month, 170 of the 290 responses indicated they felt the economy was expanding. At the .05 significance level, can we conclude that a larger proportion of the agents believe the economy is expanding this month?

A recent study focused on the number of times men and women who live alone buy takeout dinners in a month. The information is summarized below. $$\begin{array}{|lcc|}\hline \text { Statistic } & \text { Men } & \text { Women } \\\\\hline \text { Mean } & 24.51 & 22.69 \\\\\text { Standard deviation } & 4.48 & 3.86 \\\\\text { Sample size } & 35 & 40 \\\\\hline\end{array}$$ At the .01 significance level, is there a difference in the mean number of times men and women order takeout dinners in a month? What is the \(p\) -value?

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