Question

find the standardized test statistic, t, to test the claim that $\mu_1 < \mu_2$. two samples are randomly selected from populations that are normal. the sample statistics are given below. assume that $\sigma_1^2 = \sigma_2^2$.$n_1 = 15, n_2 = 15, \bar{x}_1 = 22.78, \bar{x}_2 = 25.33, s_1 = 2.9, s_2 = 2.8$\\(\bigcirc\\) a. \\(-3.165\\)\\(\bigcirc\\) b. \\(-0.669\\)\\(\bigcirc\\) c. \\(-2.450\\)\\(\bigcirc\\) d. \\(-1.667\\)

Explanation:

Step1: Recall the formula for pooled t - test statistic

The formula for the standardized test statistic \(t\) for testing the claim about two population means with \(\sigma_{1}^{2}=\sigma_{2}^{2}\) is:

$$t=\frac{(\bar{x}_{1}-\bar{x}_{2})-(\mu_{1}-\mu_{2})}{\sqrt{s_{p}^{2}(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}$$

where \(s_{p}^{2}=\frac{(n_{1} - 1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2}\) and \(\mu_{1}-\mu_{2} = 0\) (since we are testing \(\mu_{1}<\mu_{2}\), the null hypothesis is \(\mu_{1}-\mu_{2}=0\)).

Step2: Calculate the pooled variance \(s_{p}^{2}\)

Given \(n_{1}=15\), \(n_{2}=15\), \(s_{1}=2.9\), \(s_{2}=2.8\)
First, calculate \((n_{1}-1)s_{1}^{2}\) and \((n_{2}-1)s_{2}^{2}\)
\((n_{1}-1)s_{1}^{2}=(15 - 1)\times(2.9)^{2}=14\times8.41 = 117.74\)
\((n_{2}-1)s_{2}^{2}=(15 - 1)\times(2.8)^{2}=14\times7.84=109.76\)
Then, \(s_{p}^{2}=\frac{117.74 + 109.76}{15 + 15-2}=\frac{227.5}{28}\approx8.125\)

Step3: Calculate the numerator of the t - statistic

\(\bar{x}_{1}=22.78\), \(\bar{x}_{2}=25.33\), \(\mu_{1}-\mu_{2}=0\)
The numerator is \((\bar{x}_{1}-\bar{x}_{2})-(\mu_{1}-\mu_{2})=(22.78 - 25.33)-0=- 2.55\)

Step4: Calculate the denominator of the t - statistic

\(\sqrt{s_{p}^{2}(\frac{1}{n_{1}}+\frac{1}{n_{2}})}=\sqrt{8.125\times(\frac{1}{15}+\frac{1}{15})}=\sqrt{8.125\times\frac{2}{15}}=\sqrt{\frac{16.25}{15}}\approx\sqrt{1.0833}\approx1.0408\)

Step5: Calculate the t - statistic

\(t=\frac{-2.55}{1.0408}\approx - 2.45\) (rounded to three decimal places)

Answer:

C. - 2.450