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Chapter 10: Problem 33

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. $$ \begin{array}{lll} n_{1}=14 & \bar{x}_{1}=109.43 & s_{1}=2.26 \\ n_{2}=15 & \bar{x}_{2}=113.88 & s_{2}=5.84 \end{array} $$

Short Answer

Expert verified
The \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) is \((-8.21, -0.69)\)

Step by step solution

01

Compute sample mean difference and standard errors

Firstly, we need to compute the difference of the sample means \(\bar{x}_1 - \bar{x}_2\). We also need to calculate the two standard errors using the formula \(\sqrt{\frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}}}\). In the given exercise, \(\bar{x}_1 = 109.43, \bar{x}_2 = 113.88, s_1 =2.26 , s_2 = 5.84, n_1 = 14, n_2 = 15\). Therefore, \(\bar{x}_1 - \bar{x}_2 = 109.43 - 113.88 = -4.45\) and the standard error SE = \(\sqrt{\frac{2.26^2}{14} + \frac{5.84^2}{15}} = 1.796\)
02

Calculate degree of freedom

Next, to find the t score, we need to calculate the degree of freedom. For this, we use the formula for unequal variances:\(df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}}\). Plugging in the values, \(df = \frac{(\frac{2.26^2}{14} + \frac{5.84^2}{15})^2}{\frac{(\frac{2.26^2}{14})^2}{14 - 1} + \frac{(\frac{5.84^2}{15})^2}{15 - 1}} = 19.167\)
03

Find the t-score

Now we look up the t-score for \(df = 19.167\) and \(\alpha = 0.025\) (since this is a two-tailed test) in our t-distribution table. Interpolating between df = 18 and df = 21, we find t to be approximately 2.093.
04

Calculate the confidence interval

Finally, we can calculate the \(95\%\) confidence interval using the formula: \((\bar{x}_1 - \bar{x}_2) \pm (t \times SE)\). Plugging in the values from steps 1 and 3, we have: \(-4.45 \pm (2.093 \times 1.796)\), which simplifies to: \((-4.45 - 3.76, -4.45 + 3.76)\) or \((-8.21, -0.69)\).

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Most popular questions from this chapter

The following information was obtained from two independent samples selected from two normally distributed populations with unknown but equal standard deviations. $$ \begin{array}{llllllllllllll} \text { Sample 1: } & 47.7 & 46.9 & 51.9 & 34.1 & 65.8 & 61.5 & 50.2 & 40.8 & 53.1 & 46.1 & 47.9 & 45.7 & 49.0 \\ \text { Sample 2: } & 50.0 & 47.4 & 32.7 & 48.8 & 54.0 & 46.3 & 42.5 & 40.8 & 39.0 & 68.2 & 48.5 & 41.8 & \end{array} $$ a. Let \(\mu_{1}\) be the mean of population 1 and \(\mu_{2}\) be the mean of population \(2 .\) What is the point estimate of \(\mu_{1}-\mu_{2} ?\) b. Construct a \(98 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). c. Test at the \(1 \%\) significance level if \(\mu_{1}\) is greater than \(\mu_{2}\).

The U.S. Department of Labor collects data on unemployment insurance payments. Suppose that during 2009 a random sample of 70 unemployed people in Alabama received an average weekly benefit of \(\$ 199.65\), whereas a random sample of 65 unemployed people in Mississippi received an average weekly benefit of \(\$ 187.93\). Assume that the population standard deviations of all weekly unemployment benefits in Alabama and Mississippi are \(\$ 32.48\) and \(\$ 26.15\), respectively. a. Let \(\mu_{1}\) and \(\mu_{2}\) be the means of all weekly unemployment benefits in Alabama and Mississippi paid during 2009 , respectively. What is the point estimate of \(\mu_{1}-\mu_{2}\) ? b. Construct a \(96 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). c. Using the \(4 \%\) significance level, can you conclude that the means of all weekly unemployment benefits in Alabama and Mississippi paid during 2009 are different? Use both approaches to make this test.

The following information was obtained from two independent samples selected from two populations with unknown but equal standard deviations. $$ \begin{array}{lll} n_{1}=55 & \bar{x}_{1}=90.40 & s_{1}=11.60 \\ n_{2}=50 & \bar{x}_{2}=86.30 & s_{2}=10.25 \end{array} $$ a. What is the point estimate of \(\mu_{1}-\mu_{2}\) ? b. Construct a \(99 \%\) confidence interval for \(\mu_{1}-\mu_{2}\).

A company that has many department stores in the southern states wanted to find at two such stores the percentage of sales for which at least one of the items was returned. A sample of 800 sales randomly selected from Store A showed that for 280 of them at least one item was returned. Another sample of 900 sales randomly selected from Store B showed that for 279 of them at least one item was returned. a. Construct a \(98 \%\) confidence interval for the difference between the proportions of all sales at the two stores for which at least one item is returned. b. Using the \(1 \%\) significance level, can you conclude that the proportions of all sales for which at least one item is returned is higher for Store A than for Store \(B\) ?

A study in the July 7,2009 , issue of \(U S A\) TODAY stated that the \(401(\mathrm{k})\) participation rate among U.S. employees of Asian heritage is \(76 \%\), whereas the participation rate among U.S. employees of Hispanic heritage is \(66 \%\). Suppose that these results were based on random samples of 100 U.S. employees from each group. a. Construct a \(95 \%\) confidence interval for the difference between the two population proportions. b. Using the \(5 \%\) significance level, can you conclude that the \(401(\mathrm{k})\) participation rates are different for all U.S. employees of Asian heritage and all U.S. employees of Hispanic heritage? Use the critical- value and \(p\) -value approaches. c. Repeat parts a and b for both sample sizes of 200 instead of 100 . Does your conclusion change in part b?
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