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Chapter 10: Problem 2

Consider the following hypothesis test. $$\begin{array}{l} H_{0}: \mu_{1}-\mu_{2} \leq 0 \\ H_{\mathrm{a}}: \mu_{1}-\mu_{2}>0 \end{array}$$ The following results are for two independent 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. Test statistic \( z \approx 2.03 \). b. p-value \( \approx 0.0212 \). c. Reject \( H_0 \); conclude \( \mu_1 > \mu_2 \) at \( \alpha = 0.05 \).

Step by step solution

01

Formulate the Null and Alternative Hypotheses

For this exercise, we are testing whether the mean of population 1, \( \mu_1 \), is greater than the mean of population 2, \( \mu_2 \). The null hypothesis \( H_0 \) is that the difference between the two means is less than or equal to zero, i.e., \( H_{0}: \mu_{1}-\mu_{2} \leq 0 \). The alternative hypothesis \( H_a \) is that this difference is greater than zero, i.e., \( H_{a}: \mu_{1}-\mu_{2} > 0 \).
02

Calculate the Test Statistic

We use the formula for the test statistic for two independent sample means, given by:\[ z = \frac{(\bar{x}_1 - \bar{x}_2) - D_0}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} \]where \( D_0 = 0 \) (the hypothesized difference under \( H_0 \)). Plugging in the given values:\[ z = \frac{(25.2 - 22.8) - 0}{\sqrt{\frac{5.2^2}{40} + \frac{6.0^2}{50}}} \]\[ z = \frac{2.4}{\sqrt{0.676 + 0.72}} \]\[ z = \frac{2.4}{\sqrt{1.396}} \]\[ z = \frac{2.4}{1.1815} \] \[ z \approx 2.03 \]
03

Determine the p-value

The test statistic \( z = 2.03 \) indicates the number of standard deviations the sample mean difference is from the hypothesized difference of zero in the direction of the alternative hypothesis. Using a standard normal distribution table or calculator for a one-tailed test, find the p-value corresponding to \( z = 2.03 \). It comes out to approximately 0.0212.
04

Make a Hypothesis Testing Conclusion

Compare the p-value with the significance level \( \alpha = 0.05 \). Since the p-value (0.0212) is less than \( \alpha \), we reject the null hypothesis \( H_0 \). Thus, there is sufficient evidence to conclude that the mean of population 1 is greater than the mean of population 2 (\( \mu_1 > \mu_2 \)).

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Key Concepts

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

Test Statistic Calculation
Calculating the test statistic is an essential part of conducting a hypothesis test. In cases with independent samples, this calculation helps us figure out how far our observed sample statistics deviate from the null hypothesis.

For our data, where we have two independent samples, the formula used is:
  • \[ z = \frac{(\bar{x}_1 - \bar{x}_2) - D_0}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} \]
Where:
  • \( \bar{x}_1 \) and \( \bar{x}_2 \) are the sample means.
  • \( \sigma_1^2 \) and \( \sigma_2^2 \) are the variances of the two samples.
  • \( n_1 \) and \( n_2 \) are the sample sizes.
  • \( D_0 \) is the hypothesized population mean difference, typically 0 when testing for equality.
The test statistic we calculated turned out to be approximately 2.03, indicating how many standard deviations away the sample mean difference is from zero.
P-Value Interpretation
The p-value helps in interpreting the results of your hypothesis test by quantifying how likely it is to observe data at least as extreme as your sample, assuming that the null hypothesis is true.
In simpler terms, it tells us the probability of finding the observed, or more extreme, outcomes. For a test statistic of 2.03, we find a p-value of about 0.0212.

When considering p-value:
  • A small p-value (typically 鈮� 0.05) indicates strong evidence against the null hypothesis, so you reject the null hypothesis.
  • A large p-value (> 0.05) means weak evidence against the null hypothesis, so you fail to reject the null hypothesis.
In our exercise, given that our p-value is 0.0212, which is less than the typical significance level \( \alpha = 0.05 \), we conclude there's enough evidence to reject the null hypothesis.
Independent Samples
Independent samples mean that the two groups being compared originate from separate populations, with no influence on each other. This characteristic is crucial for certain types of hypothesis tests, like the one we are dealing with here.

Understanding independence is critical because:
  • It ensures that there's no overlap or interaction between the groups.
  • It validates the use of specific formulas and techniques in hypothesis testing, such as the test statistic calculation used above.
In this exercise, each sample's outcomes are entirely independent, making this appropriate for a two-sample z-test. Recognizing the samples are independent reassures us that our calculated results accurately reflect the difference (or lack thereof) between the two population means.

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Most popular questions from this chapter

Airline travelers often choose which airport to fly from based on flight cost. cost data (in dollars) for a sample of flights to eight cities from Dayton, Ohio, and Louisville, Kentucky, were collected to help determine which of the two airports was more costly to fly from (The Cincinnati Enquirer; February 19,2006 ). A researcher argued that it is significantly more costly to fly out of Dayton than Louisville. Use the sample data to see whether they support the researcher's argument. Use \(\alpha=.05\) as the level of significance. $$\begin{array}{lcr} \text { Destination } & \text { Dayton } & \text { Louisville } \\ \text { Chicago-O'Hare } & \$ 319 & \$ 142 \\ \text { Grand Rapids, Michigan } & 192 & 213 \\ \text { Portland, Oregon } & 503 & 317 \\ \text { Atlanta } & 256 & 387 \\ \text { Seattle } & 339 & 317 \\ \text { South Bend, Indiana } & 379 & 167 \\ \text { Miami } & 268 & 273 \\ \text { Dallas-Ft. Worth } & 288 & 274 \end{array}$$

A well-known automotive magazine took three top-of-the-line midsize automobiles manufactured in the United States, test-drove them, and compared them on a variety of criteria. In the area of gasoline mileage performance, five automobiles of each brand were each test-driven 500 miles; the miles per gallon data obtained follow. Use \(\alpha=.05\) to test whether there is a significant difference in the mean number of miles per gallon for the three types of automobiles. $$\begin{array}{ccc} & \text { Automobile } & \\ \mathbf{A} & \mathbf{B} & \mathbf{C} \\ 19 & 19 & 24 \\ 21 & 20 & 26 \\ 20 & 22 & 23 \\ 19 & 21 & 25 \\ 21 & 23 & 27 \end{array}$$

Injuries to Major League Baseball players have been increasing in recent years. For the period 1992 to 2001 , league expansion caused Major League Baseball rosters to increase \(15 \%\) However, the number of players being put on the disabled list due to injury increased \(32 \%\) over the same period (USA Today, July 8,2002 ). A research question addressed whether Major League Baseball players being put on the disabled list are on the list longer in 2001 than players put on the disabled list a decade earlier.a. Using the population mean number of days a player is on the disabled list, formulate null and alternative hypotheses that can be used to test the research question. b. Assume that the following data apply: \(\begin{array}{lll} & \mathbf{2 0 0 1} \text { Season } & \mathbf{1 9 9 2} \text { Season } \\ \text { Sample size } & n_{1}=45 & n_{2}=38 \\ \text { Sample mean } & \bar{x}_{1}=60 \text { days } & \bar{x}_{2}=51 \text { days } \\\ \text { Sample standard deviation } & s_{1}=18 \text { days } & s_{2}=15 \text { days }\end{array}\) What is the point estimate of the difference between population mean number of days on the disabled list for 2001 compared to \(1992 ?\) What is the percentage increase in the number of days on the disabled list? c. Use \(\alpha=.01 .\) What is your conclusion about the number of days on the disabled list? What is the \(p\) -value? d. Do these data suggest that Major League Baseball should be concerned about the situation?

In a study conducted to investigate browsing activity by shoppers, each shopper was initially classified as a nonbrowser, light browser, or heavy browser. For each shopper, the study obtained a measure to determine how comfortable the shopper was in a store. Higher scores indicated greater comfort. Suppose the following data were collected. Use \(\alpha=.05\) to test for differences among comfort levels for the three types of browsers. $$\begin{array}{ccc} & \text { Light } & \text { Heavy } \\ \text { Nonbrowser } & \text { Browser } & \text { Browser } \\ 4 & 5 & 5 \\ 5 & 6 & 7 \\ 6 & 5 & 5 \\ 3 & 4 & 7 \\ 3 & 7 & 4 \\ 4 & 4 & 6 \\ 5 & 6 & 5 \\ 4 & 5 & 7 \end{array}$$

To study the effect of temperature on yield in a chemical process, five batches were produced at each of three temperature levels. The results follow. Construct an analysis of variance table. Use a .05 level of significance to test whether the temperature level has an effect on the mean yield of the process. $$\begin{array}{ccc} & \text { Temperature } & \\ \mathbf{5 0}^{\circ} \mathbf{C} & \mathbf{6 0}^{\circ} \mathbf{C} & \mathbf{7 0}^{\circ} \mathbf{C} \\ 34 & 30 & 23 \\ 24 & 31 & 28 \\ 36 & 34 & 28 \\ 39 & 23 & 30 \\ 32 & 27 & 31 \end{array}$$
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