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The following information was obtained from two independent samples selected from two populations with unknown but equal standard deviations. $$ \begin{array}{lll} n_{1}=55 & \bar{x}_{1}=90.40 & s_{1}=11.60 \\ n_{2}=50 & \bar{x}_{1}=86.30 & s_{2}=10.25 \end{array} $$ Test at a \(5 \%\) significance level if \(\mu_{1}\) is greater than \(\mu_{2}\).

Short Answer

Expert verified
The final decision will depend on whether the calculated test statistic is greater than the critical value. If it's greater, then it can be concluded with a \(95%\) confidence that \(\mu_{1}\) is greater than \(\mu_{2}\). If not, there's no statistical evidence to prove that \(\mu_{1}\) is greater than \(\mu_{2}\).

Step by step solution

01

State the hypotheses.

The null hypothesis \( H_{0}: \mu_{1} = \mu_{2} \) assumes that the means of the two populations are equal. The alternative hypothesis \( H_{1}: \mu_{1} > \mu_{2} \) assumes that the population mean \(\mu_{1}\) is greater than \(\mu_{2}\).
02

Determine the level of significance.

The level of significance, also the probability of rejecting the null hypothesis when it is true, is typically denoted by \(\alpha\). Here, \(\alpha = 0.05 \) (or \(5%\)).
03

Calculate the pooled variance.

The pooled variance is an estimate of the common within-group variance computed by taking the weighted average of the within-group variances. The formula used is \( s_{p}^{2} = \frac{(n_1-1)s_{1}^{2} + (n_2-1)s_{2}^{2}}{n_1+n_2-2} \) where \( n_1 \) and \( n_2 \) are the sample sizes, and \( s_{1}^{2} \) and \( s_{2}^{2} \) are the variances of the two samples. Plug in the given values to calculate the pooled variance.
04

Compute the Test Statistic.

The test statistic for comparing two means is given by \( T = \frac{\bar{x}_{1}-\bar{x}_{2}}{s_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}} \). Using the computed pooled variance and the given means and sample sizes, the test statistic can be calculated.
05

Find the Critical Value.

The critical value is the value that the test statistic must exceed in order for the null hypothesis to be rejected. The critical value can be found using the t-distribution table, with df = \( n_1 + n_2 - 2 \) and \( \alpha \)
06

Decision-making.

If the test statistic is greater than the critical value, reject the null hypothesis; hence, the alternative hypothesis is accepted that is, \(\mu_{1}\) is greater than \(\mu_{2}\). If not, we fail to reject the null hypothesis; hence, it's not statistically significant that \(\mu_{1}\) is greater than \(\mu_{2}\).

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Key Concepts

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

Pooled Variance
When comparing two sample groups, it’s essential to estimate the variance shared between the groups. This is where the concept of pooled variance comes into play. Pooled variance is useful when the standard deviations for both populations are believed to be equal. It provides a more precise measure by combining individual sample variances, giving us a weighted average.
\[ s_{p}^{2} = \frac{(n_1-1)s_{1}^{2} + (n_2-1)s_{2}^{2}}{n_1+n_2-2} \]
Here, \(n_1\) and \(n_2\) are sample sizes, while \(s_{1}^{2}\) and \(s_{2}^{2}\) depict the sample variances. The pooled variance then becomes necessary for further calculations such as the test statistic. By using this formula, you calculate a pooled variance, which helps in determining if there is a statistically significant difference in means between the two groups.
Significance Level
In hypothesis testing, the significance level, denoted as \(\alpha\), is a crucial concept. It represents the probability of rejecting the null hypothesis when it is actually true. This is what we call a Type I error. A common choice for \(\alpha\) is 0.05, which corresponds to a 5% risk of making such an error.
  • A lower significance level means a stricter criterion for rejecting the null hypothesis.
  • Setting \(\alpha = 0.05\) means you are 95% confident in your results.
In this problem, we're using a 5% significance level, indicating we are willing to accept a 5% chance of concluding that there is a difference in means when, in fact, there is none. Choosing an appropriate significance level is vital for the validity and reliability of any test.
Critical Value
The critical value is a threshold that the test statistic must exceed to reject the null hypothesis. It helps determine the significance of test results. You typically find it from a statistical table, such as the t-distribution table.
  • To find the critical value, you need the degrees of freedom, calculated as \(df = n_1 + n_2 - 2\).
  • Using \(\alpha = 0.05\), find the value on the t-table that corresponds to your degrees of freedom.
In simpler terms, the critical value is the borderline between accepting or rejecting the null hypothesis based on your test statistic. If your test statistic exceeds this critical value, the results are deemed statistically significant.
Test Statistic
The test statistic is a standardized value that is calculated from sample data. It is used to decide whether to reject the null hypothesis. In tests comparing two means with equal variances, the t-statistic is utilized.
\[ T = \frac{\bar{x}_{1}-\bar{x}_{2}}{s_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}} \] \(\bar{x}_{1}\) and \(\bar{x}_{2}\) stand for sample means, while \(s_p\) denotes pooled variance. This formula essentially measures how many standard errors the observed difference between sample means is away from zero.
  • The higher the test statistic, the further the difference from what would be expected under the null hypothesis.
  • It provides the basis for decision-making regarding whether to accept or reject the null hypothesis.
Once calculated, you compare the test statistic to the critical value to check for statistical significance.

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Most popular questions from this chapter

In a survey of American drivers, \(79 \%\) of women drivers and \(85 \%\) of men drivers said that they exceeded the speed limit at least once in the past week. Suppose that these percentages are based on random samples of 600 women and 700 men drivers. a. Let \(p_{1}\) and \(p_{2}\) be the proportion of all women and men American drivers, respectively, who will say that they exceeded the speed limit at least once in the past week. Construct a \(98 \%\) confidence interval for \(p_{1}-p_{2}\). b. Using a \(1 \%\) significance level, can you conclude that \(p_{1}\) is lower than \(p_{2}\) ? Use both the critical-value and the \(p\) -value approaches.

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.

Sample of 500 observations taken from the first population gave \(x_{1}=305\). Another sample of 600 observations taken from the second population gave \(x_{2}=348\). a. Find the point estimate of \(p_{1}-p_{2}\). b. Make a \(97 \%\) confidence interval for \(p_{1}-p_{2}\). c. Show the rejection and nonrejection regions on the sampling distribution of \(\hat{p}_{1}-\hat{p}_{2}\) for \(H_{0}: p_{1}=p_{2}\) versus \(H_{1}: p_{1}>p_{2}\). Use a significance level of \(2.5 \%\). d. Find the value of the test statistic \(z\) for the test of part \(\mathrm{c}\). e. Will you reject the null hypothesis mentioned in part \(\mathrm{c}\) at \(\mathrm{a}\) significance level of \(2.5 \% ?\)

A car magazine is comparing the total repair costs incurred during the first three years on two sports cars, the T-999 and the XPY. Random samples of 45 T-999s and 51 XPYs are taken. All 96 cars are 3 years old and have similar mileages. The mean of repair costs for the 45 T-999 cars is \(\$ 3300\) for the first 3 years. For the 51 XPY cars, this mean is \(\$ 3850\). Assume that the standard deviations for the two populations are \(\$ 800\) and \(\$ 1000\), respectively. a. Construct a 99\% confidence interval for the difference between the two population means. b. Using a \(1 \%\) significance level, can you conclude that such mean repair costs are different for these two types of cars? c. What would your decision be in part b if the probability of making a Type I error were zero? Explain.

Assuming that the two populations are normally distributed with unequal and unknown population standard deviations, construct a \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) for the following. $$ \begin{array}{lll} n_{1}=14 & \bar{x}_{1}=109.43 & s_{1}=2.26 \\ n_{2}=15 & \bar{x}_{2}=113.88 & s_{2}=5.84 \end{array} $$

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