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Chapter 11: Problem 3

Perform the appropriate hypothesis test. A random sample of size \(n=13\) obtained from a population that is normally distributed results in a sample mean of 45.3 and sample standard deviation of \(12.4 .\) An independent sample of size \(n=18\) obtained from a population that is normally distributed results in a sample mean of 52.1 and sample standard deviation of 14.7 . Does this constitute sufficient evidence to conclude that the population means differ at the \(\alpha=0.05\) level of significance?

Short Answer

Expert verified
No, there is not sufficient evidence to conclude that the population means differ at the \( \alpha = 0.05 \) level of significance.

Step by step solution

01

Formulate the Hypotheses

Determine the null and alternative hypotheses. The null hypothesis \( H_0 \) states that there is no difference in population means, i.e., \( \mu_1 = \mu_2 \). The alternative hypothesis \( H_a \) states that the population means are different, i.e., \( \mu_1 \eq \mu_2 \).
02

Determine the Test Statistic

Since the population standard deviations are unknown and the samples are independent and normally distributed, use the t-test for comparing two means. The test statistic is calculated as: \[ t = \frac{(\bar{x}_1 - \bar{x}_2)}{\sqrt{\frac{s^2_1}{n_1} + \frac{s^2_2}{n_2}}} \] where \( \bar{x}_1 = 45.3 \), \( \bar{x}_2 = 52.1 \), \( s_1 = 12.4 \), \( s_2 = 14.7 \), \( n_1 = 13 \), and \( n_2 = 18 \).
03

Calculate the Test Statistic

Plug in the values to calculate the test statistic: \[ t = \frac{(45.3 - 52.1)}{\sqrt{\frac{12.4^2}{13} + \frac{14.7^2}{18}}} \] Simplify the expression: \[ t = \frac{-6.8}{\sqrt{\frac{153.76}{13} + \frac{216.09}{18}}} = \frac{-6.8}{\sqrt{11.82 + 12.004}} = \frac{-6.8}{4.857} \approx -1.40 \]
04

Determine the Degrees of Freedom

The degrees of freedom for the test can be approximated using the formula: \[ df = \frac{\left(\frac{s^2_1}{n_1} + \frac{s^2_2}{n_2}\right)^2}{\frac{\left(\frac{s^2_1}{n_1}\right)^2}{n_1-1} + \frac{\left(\frac{s^2_2}{n_2}\right)^2}{n_2-1}} \] Substituting the given values: \[ df = \frac{(\frac{153.76}{13} + \frac{216.09}{18})^2}{\frac{\left(\frac{153.76}{13}\right)^2}{12} + \frac{\left(\frac{216.09}{18}\right)^2}{17}} \approx 28.275 \] Use the nearest whole number, so \( df \approx 28 \).
05

Determine the Critical Value

For a two-tailed test at \( \alpha = 0.05 \) and \( df = 28 \), the critical value from the t-distribution table is approximately \( \pm 2.048 \).
06

Compare the Test Statistic and Critical Value

The test statistic calculated is \( t = -1.40 \) and the critical values are \( \pm 2.048 \). Since \( -2.048 < -1.40 < 2.048 \, \, \text{we fail to reject the null hypothesis}. \)
07

State the Conclusion

Since we fail to reject the null hypothesis, there is not sufficient evidence to conclude that the population means differ at the \( \alpha = 0.05 \) level of significance.

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Key Concepts

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

t-test for two means
A t-test for two means helps determine if there is a significant difference between the means of two populations. This test is used when population standard deviations are unknown and the samples are independent and normally distributed. The test statistic is: } , {
statistical significance
Statistical significance tells us if our results are likely to have occurred by chance. In hypothesis testing, we often use a significance level (commonly denoted as \(\beta\)) to compare against our computed p-value. In this exercise, \(\beta = 0.05\), meaning we are 95% confident.
degrees of freedom
Degrees of freedom (df) dictate the shape of the t-distribution. They are calculated based on the sample size and are crucial for identifying critical values. In the exercise, df were calculated based on the size and variability of two populations, giving us df = 28.
critical value
The critical value is a threshold we use to compare our test statistic. It's determined by the significance level (a) and degrees of freedom (df). In this exercise, for % alpha = 0.05, the critical value for our two-tailed test with df = 28 is +-2.048.
population means comparison
To compare populations' means, we test if the difference observed from samples could be due to random error. In this problem, the hypothesis test showed that the sample means BO 45.3 and 52.1 were not sufficiently distinct to conclude population means differ at 伪 = 0.05 significance level.

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Most popular questions from this chapter

Clifford Adelman, a researcher with the Department of Education, followed a cohort of students who graduated from high school in \(1992,\) monitoring the progress the students made toward completing a bachelor's degree. One aspect of his research was to compare students who first attended a community college to those who immediately attended and remained at a four-year institution. The sample standard deviation time to complete a bachelor's degree of the 268 students who transferred to a four-year school after attending community college was \(1.162 .\) The sample standard deviation time to complete a bachelor's degree of the 1145 students who immediately attended and remained at a four-year institution was \(1.015 .\) Assuming the time to earn a bachelor's degree is normally distributed, does the evidence suggest the standard deviation time to earn a bachelor's degree is different between the two groups? Use the \(\alpha=0.05\) level of significance.

Explain the difference between an independent and dependent sample.

For each study, explain which statistical procedure (estimating a single proportion; estimating a single mean; hypothesis test for a single proportion; hypothesis test for a single mean; hypothesis test or estimation of two proportions, hypothesis test or estimation of two means, dependent or independent) would most likely be used for the research objective given. Assume all model requirements for conducting the appropriate procedure have been satisfied. Do adult males who take a single aspirin daily experience a lower rate of heart attacks than adult males who do not take aspirin daily?

What is the typical age difference between husband and wife? The following data represent the ages of husbands and wives, based on results from the Current Population Survey. (a) What is the response variable in this study? (b) Is the sampling method dependent or independent? Explain. (c) Use the data to estimate the mean difference in age of husband and wives with \(95 \%\) confidence. Explain the technique that you used.

Walking in the Airport, Part I Do people walk faster in the airport when they are departing (getting on a plane) or when they are arriving (getting off a plane)? Researcher Seth B. Young measured the walking speed of travelers in San Francisco International Airport and Cleveland Hopkins International Airport. His findings are summarized in the table. $$ \begin{array}{lcc} \text { Direction of Travel } & \text { Departure } & \text { Arrival } \\ \hline \text { Mean speed (feet per minute) } & 260 & 269 \\ \hline \begin{array}{l} \text { Standard deviation } \\ \text { (feet per minute) } \end{array} & 53 & 34 \\ \hline \text { Sample size } & 35 & 35 \\ \hline \end{array} $$ (a) Is this an observational study or a designed experiment? Why? (b) Explain why it is reasonable to use Welch's \(t\) -test. (c) Do individuals walk at different speeds depending on whether they are departing or arriving at the \(\alpha=0.05\) level of significance?
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