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Chapter 10: Problem 2

Consider the following hypothesis test. \(H_{0}: \mu_{1}-\mu_{2} \leq 0\) \(H_{\mathrm{a}}: \mu_{1}-\mu_{2}>0\) The following results are for two independent random samples taken from the two populations. \(\begin{array}{ll}\text { Sample } 1 & \text { Sample } 2 \\\ n_{1}=40 & n_{2}=50 \\ \bar{x}_{1}=25.2 & \bar{x}_{2}=22.8 \\ \sigma_{1}=5.2 & \sigma_{2}=6.0\end{array}\) a. What is the value of the test statistic? b. What is the \(p\) -value? c. With \(\alpha=.05,\) what is your hypothesis testing conclusion?

Short Answer

Expert verified
a. The test statistic is approximately 2.03. b. The p-value is approximately 0.021. c. Reject the null hypothesis; there is evidence that \( \mu_1 > \mu_2 \).

Step by step solution

01

Understand the Hypotheses

Identify the null and alternative hypotheses. In this case, the null hypothesis is \( H_0: \mu_1 - \mu_2 \leq 0 \) and the alternative hypothesis is \( H_a: \mu_1 - \mu_2 > 0 \).
02

Find Test Statistic Formula

For two independent samples with known standard deviations, use the formula for the z-test statistic: \( z = \frac{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} \). Here, \( \mu_1 - \mu_2 = 0 \) under the null.
03

Calculate Standard Error

Calculate the standard error for the difference of means: \[ \text{SE} = \sqrt{\frac{5.2^2}{40} + \frac{6.0^2}{50}} = \sqrt{\frac{27.04}{40} + \frac{36}{50}} = \sqrt{0.676 + 0.72} = \sqrt{1.396} \approx 1.18 \].
04

Compute the Test Statistic

Substitute the values into the z-test formula: \( z = \frac{25.2 - 22.8}{1.18} = \frac{2.4}{1.18} \approx 2.03 \).
05

Find the p-value

Use the z-value to find the p-value from standard z-tables for a one-tailed test. For \( z = 2.03 \), \( p \approx 0.021 \) (look up the cumulative probability and subtract from 1).
06

Compare p-value with Alpha

Compare the p-value (0.021) with the significance level \( \alpha = 0.05 \). Since the p-value is less than \( \alpha \), reject the null hypothesis.
07

Conclusion

Based on the comparison, you conclude that there is enough evidence to support the claim that \( \mu_1 - \mu_2 > 0 \) at the \( \alpha = 0.05 \) significance level.

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Key Concepts

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

z-test
The z-test is a statistical tool used when comparing the means from two different populations to see if there's any significant difference. It's particularly handy when the standard deviations of the populations are known. The test helps in knowing if any observed difference is just by chance or if it鈥檚 significant enough to draw conclusions.
To perform a z-test, you calculate the z-statistic using the formula:
  • If the z-value is far from 0, it suggests a significant difference between sample means.
  • A z-statistic is computed by dividing the difference between the sample means by the standard error (SE).
The larger the z-statistic, the stronger the evidence against the null hypothesis, indicating a significant difference between the means.
p-value
The p-value is an essential component in hypothesis testing, as it helps us determine the significance of our results. It represents the probability of observing results as extreme as the ones seen in your data, assuming the null hypothesis is true.
For a given z-test:
  • A smaller p-value indicates stronger evidence against the null hypothesis.
  • In our example, a p-value of 0.021 suggests that observing a difference as large as the suspected one between the two population means is very unlikely if the null hypothesis is true.
Whenever the p-value is less than the significance level (usually 0.05), you reject the null hypothesis, concluding that there is a statistically significant difference.
significance level
The significance level, denoted by \( \alpha \), is a threshold set before conducting any statistical test. It defines the probability of rejecting the null hypothesis when it's actually true, which is known as a Type I error. Commonly, researchers use a significance level of 0.05.
In hypothesis testing:
  • If your p-value is less than or equal to \( \alpha \), reject the null hypothesis.
  • The chosen \( \alpha = 0.05 \) gives a 5% risk of incorrectly rejecting a true null hypothesis.
This means you are 95% confident in your decision when rejecting the null hypothesis, providing a balance between risk and accuracy.
standard error
The standard error (SE) plays a crucial role in hypothesis testing by estimating the variability of sample means around the population mean. It's used to measure how much we expect a sample mean to vary from the true population mean.
The formula for the standard error when comparing two independent samples is: \[ \text{SE} = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}} \]
  • A small SE implies that the sample mean is a precise estimate of the population mean.
  • In our example, SE is calculated to be approximately 1.18.
Accurate calculation and understanding of SE is crucial, as it directly influences the z-statistic and ultimately the hypothesis test result.

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Most popular questions from this chapter

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The following data are from a completely randomized design. \(\begin{array}{ccc} & \text { Treatment } & \\ \mathbf{A} & \mathbf{B} & \mathbf{C} \\ 162 & 142 & 126 \\ 142 & 156 & 122 \\ 165 & 124 & 138 \\ 145 & 142 & 140 \\ 148 & 136 & 150 \\ 174 & 152 & 128\end{array}\) \(\begin{array}{lccc}\text { Sample mean } & 156 & 142 & 134 \\ \text { Sample variance } & 164.4 & 131.2 & 110.4\end{array}\) a. Compute the sum of squares between treatments. b. Compute the mean square between treatments. c. Compute the sum of squares due to error d. Compute the mean square due to error. e. Set up the ANOVA table for this problem. f. \(\quad\) At the \(\alpha=.05\) level of significance, test whether the means for the three treatments are equal.

Auditors must make judgments about various aspects of an audit on the basis of their own direct experience, indirect experience, or a combination of the two. In a study, auditors were asked to make judgments about the frequency of errors to be found in an audit. The judgments by the auditors were then compared to the actual results. Suppose the following data were obtained from a similar study; lower scores indicate better judgments. \(\begin{array}{ccc}\text { Direct } & \text { Indirect } & \text { Combination } \\ 17.0 & 16.6 & 25.2 \\ 18.5 & 22.2 & 24.0 \\ 15.8 & 20.5 & 21.5 \\\ 18.2 & 18.3 & 26.8 \\ 20.2 & 24.2 & 27.5 \\ 16.0 & 19.8 & 25.8 \\ 13.3 & 21.2 & 24.2\end{array}\) Use \(\alpha=.05\) to test to see whether the basis for the judgment affects the quality of the judgis
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