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The following information was obtained from two independent samples selected from two normally distributed populations with unknown but equal standard deviations. $$ \begin{array}{lll} n_{1}=21 & \bar{x}_{1}=13.97 & s_{1}=3.78 \\ n_{2}=20 & \bar{x}_{2}=15.55 & s_{2}=3.26 \end{array} $$ Test at a \(5 \%\) significance level if the two population means are different.

Short Answer

Expert verified
After carrying out the above steps, if the test statistic T is found to be greater than the upper critical value or less than the lower one, we can conclude that there is enough evidence to reject the null hypothesis, so the means are different. But, if T is between the critical values, there is not enough evidence to reject the null hypothesis, so the means are not significantly different.

Step by step solution

01

Set up the hypotheses

The null hypothesis (H0) and the alternative hypothesis (H1) are as follows: \n \nH0: \(\mu_{1} = \mu_{2}\) - The population means are equal. \n\nH1: \(\mu_{1} \neq \mu_{2}\) - The population means are not equal.
02

Calculate the pooled standard deviation

The pooled standard deviation \(s_{p}\) is calculated as follows: \n\n\(s_{p} = \sqrt{\frac{(n_{1}-1)s_{1}^{2} + (n_{2}-1)s_{2}^{2}} {n_{1}+n_{2}-2}}\)\n\nSubstitute the given values: \n\ \(n_{1} = 21, s_{1} = 3.78, n_{2} = 20, s_{2} = 3.26\) \n\nThen, calculate to find \(s_{p}\).
03

Calculate the Test Statistic

The test statistic (T) is calculated as follows: \n\nT = \(\frac{(\bar{x}_{1} - \bar{x}_{2})- ( \mu_{1} - \mu_{2})}{sp\sqrt{\frac{1}{n_{1}} + \frac{1}{n_{2}}}}\)\n\nSince H0 posits that \(\mu_{1} - \mu_{2} = 0\), the equation simplifies to: \n\n T = \(\frac{(\bar{x}_{1} - \bar{x}_{2})}{sp\sqrt{\frac{1}{n_{1}} + \frac{1}{n_{2}}}}\)\n\nSubstitute the given values: \n\(\bar{x}_{1} = 13.97, \bar{x}_{2} = 15.55, n_{1} = 21, n_{2}=20, s_{p}\) (from Step 2). \n\nThen compute for T.
04

Determine the Critical Value

Go to the t-distribution table given a 5% significance level and degrees of freedom (DF) {DF = \(n_{1} + n_{2} - 2\)} to find the critical value. \n\nCompare the test statistic to the critical values. If the test statistic is beyond the critical values (less than the lower critical value or greater than the upper one), reject the null hypothesis. If it is between the two critical values, do not reject the null hypothesis.

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Key Concepts

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

t-distribution
In hypothesis testing, when dealing with small sample sizes or unknown population standard deviations, we often turn to the **t-distribution**. Unlike the Z-distribution, the t-distribution is more spread out, especially with smaller sample sizes. This distribution becomes narrower as the sample size increases, eventually resembling the normal distribution for large samples.

The use of the t-distribution is crucial when comparing sample means, especially when the population standard deviation is not known. In this exercise, since the sample sizes are relatively small (21 and 20), and the population standard deviations are unknown, the t-distribution is appropriate. We also consider the degrees of freedom to be \( n_1 + n_2 - 2 \) when calculating the critical value from the t-table.

This choice ensures that our hypothesis test is as accurate as possible given the sample constraints. Understanding when and how to apply the t-distribution is a foundational concept in statistics, especially in inferential statistics.
significance level
The **significance level**, often denoted by \( \alpha \), is a threshold we use in hypothesis testing to decide whether to reject the null hypothesis. A common significance level is 5%, or \( 0.05 \). This level indicates the probability of rejecting the null hypothesis when it is actually true, which is known as a Type I error.

In our test, we are using a 5% significance level. This means we are willing to accept a 5% risk of incorrectly claiming that the population means are different when they are actually equal. The significance level directly affects the critical values drawn from the t-distribution; these values determine the threshold for rejecting the null hypothesis.

By setting the significance level before conducting the test, we establish clear criteria for decision-making, allowing us to make informed conclusions based on statistical evidence rather than random chance.
pooled standard deviation
The **pooled standard deviation** is a way to estimate a common standard deviation across two independent samples. In cases where the samples are taken from populations with unknown but assumed equal variances, pooling their variances provides a more precise estimate.

The formula to calculate the pooled standard deviation \( s_p \) is:
\[s_p = \sqrt{\frac{(n_{1}-1)s_{1}^{2} + (n_{2}-1)s_{2}^{2}}{n_{1} + n_{2} - 2}}\]
Where \( s_1 \) and \( s_2 \) are the sample standard deviations, and \( n_1 \) and \( n_2 \) are the sample sizes. This formula combines the variances of the two samples, weighted by their respective degrees of freedom.

Using a pooled standard deviation is essential in this context because it allows us to compare the means of two different samples while accounting for the spread of data within them. This calculation is an integral part of determining the test statistic in our hypothesis test.
population means
The concept of **population means** refers to the average values of certain characteristics of entire populations. In most statistical tests, we are interested in estimating or comparing these population means based on sample data.

In our specific problem, we are testing whether the mean from one population \( \mu_1 \) is equal to the mean from another population \( \mu_2 \). The null hypothesis \( H_0 \) states that the population means are equal (i.e., \( \mu_1 = \mu_2 \)). The alternative hypothesis \( H_1 \) posits that these means are distinct (i.e., \( \mu_1 eq \mu_2 \)).

Utilizing sample data with means \( \bar{x}_1 \) and \( \bar{x}_2 \), we make inferences about the population means. If our test statistic, derived from these sample means and standard deviations, falls outside the critical values, we may conclude that the population means are likely different. Such conclusions provide valuable insights about the differences or similarities between the studied groups.

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Most popular questions from this chapter

In parts of the eastern United States, whitetail deer are a major nuisance to farmers and homeowners, frequently damaging crops, gardens, and landscaping. A consumer organization arranges a test of two of the leading deer repellents \(\mathrm{A}\) and \(\mathrm{B}\) on the market. Fifty-six unfenced gardens in areas having high concentrations of deer are used for the test. Twenty-nine gardens are chosen at random to receive repellent A, and the other 27 receive repellent \(\mathrm{B}\). For each of the 56 gardens, the time elapsed between application of the repellent and the appearance of the first deer in the garden is recorded. For repellent \(\mathrm{A}\), the mean time is 101 hours. For repellent \(\mathrm{B}\), the mean time is 92 hours. Assume that the two populations of elapsed times have approximately normal distributions with population standard deviations of 15 and 10 hours, respectively. a. Let \(\mu_{1}\) and \(\mu_{2}\) be the population means of elapsed times for the two repellents, respectively. Find the point estimate of \(\mu_{1}-\mu_{2}\). b. Find a \(97 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). c. Test at a \(2 \%\) significance level whether the mean elapsed times for repellents \(\mathrm{A}\) and \(\mathrm{B}\) are different. Use both approaches, the critical-value and \(p\) -value, to perform this test.

A high school counselor wanted to know if tenth-graders at her high school tend to have the same free time as the twelfth-graders. She took random samples of 25 tenth-graders and 23 twelfth-graders. Each student was asked to record the amount of free time he or she had in a typical week. The mean for the tenth-graders was found to be 29 hours of free time per week with a standard deviation of \(7.0\) hours. For the twelfth-graders, the mean was 22 hours of free time per week with a standard deviation of \(6.2\) hours. Assume that the two populations are approximately normally distributed with unknown but equal standard deviations. a. Make a \(90 \%\) confidence interval for the difference between the corresponding population means. b. Test at a \(5 \%\) significance level whether the two population means are different.

The following information is obtained from two independent samples selected from two populations. $$ \begin{array}{lll} n_{1}=650 & \bar{x}_{1}=1.05 & \sigma_{1}=5.22 \\ n_{2}=675 & \bar{x}_{2}=1.54 & \sigma_{2}=6.80 \end{array} $$ a. What is the point estimate of \(\mu_{1}-\mu_{2} ?\) b. Construct a \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). Find the margin of error for this estimate.

The following information was obtained from two independent samples selected from two populations with unequal and unknown population standard deviations. $$ \begin{array}{lll} n_{1}=48 & \bar{x}_{1}=.863 & s_{1}=.176 \\ n_{2}=46 & \bar{x}_{2}=.796 & s_{2}=.068 \end{array} $$ Test at a \(2.5 \%\) significance level if \(\mu_{1}\) is greater than \(\mu_{2}\).

Consider the following information obtained from two independent samples: $$ n_{1}=300, \quad \hat{p}_{1}=.55, \quad n_{2}=200, \quad \hat{p}_{2}=.62 $$ Test at a \(1 \%\) significance level if \(p_{1}\) is less than \(p_{2}\).

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