Question

find the standardized test statistic, t, to test the claim that $\mu_1 > \mu_2$. two samples populations that are normal. the sample statistics are given below. assume that $n_1 = 18, n_2 = 13, \bar{x}_1 = 605, \bar{x}_2 = 590, s_1 = 40, s_2 = 25$ \bigcirc a. 3.271 \bigcirc b. 2.819 \bigcirc c. 1.865 \bigcirc d. 1.282

Explanation:

Step1: Recall the formula for the two - sample t - test statistic

The formula for the standardized test statistic \(t\) for testing the claim about two population means (\(\mu_1\) and \(\mu_2\)) when the populations are normal (and we assume unequal variances, which is a common case when we don't have information about equal variances) is:

\[t=\frac{(\bar{x}_1 - \bar{x}_2)-(\mu_1-\mu_2)}{\sqrt{\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2}}}\]

Since we are testing the claim that \(\mu_1>\mu_2\), the null hypothesis is \(H_0:\mu_1 - \mu_2 = 0\), so \((\mu_1-\mu_2)=0\) in the formula.

Step2: Substitute the given values into the formula

We are given \(n_1 = 18\), \(n_2=13\), \(\bar{x}_1 = 605\), \(\bar{x}_2 = 590\), \(s_1 = 40\), \(s_2=25\)

First, calculate the numerator: \(\bar{x}_1-\bar{x}_2=605 - 590=15\)

Then, calculate the denominator:

\[ \begin{align*} \sqrt{\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2}}&=\sqrt{\frac{40^{2}}{18}+\frac{25^{2}}{13}}\\ &=\sqrt{\frac{1600}{18}+\frac{625}{13}}\\ &=\sqrt{\frac{1600\times13 + 625\times18}{18\times13}}\\ &=\sqrt{\frac{20800+ 11250}{234}}\\ &=\sqrt{\frac{32050}{234}}\\ &=\sqrt{137.0085}\\ &\approx11.705 \end{align*} \]

Step3: Calculate the value of \(t\)

Now, \(t=\frac{15}{11.705}\approx1.282\)? Wait, no, wait, I made a mistake in the denominator calculation. Let's recalculate the denominator:

\(\frac{s_1^{2}}{n_1}=\frac{40^{2}}{18}=\frac{1600}{18}\approx88.8889\)

\(\frac{s_2^{2}}{n_2}=\frac{25^{2}}{13}=\frac{625}{13}\approx48.0769\)

Then \(\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2}=88.8889 + 48.0769=136.9658\)

\(\sqrt{136.9658}\approx11.703\)

Then \(t=\frac{15}{11.703}\approx1.282\)? But wait, maybe we should use the pooled variance? Wait, no, the problem doesn't state that the variances are equal. Wait, maybe I misread the sample sizes. Wait, \(n_1 = 18\), \(n_2=13\). Wait, let's check the formula again. Wait, maybe the question assumes equal variances? Let's try the pooled variance formula.

The pooled variance \(s_p^{2}=\frac{(n_1 - 1)s_1^{2}+(n_2 - 1)s_2^{2}}{n_1 + n_2-2}\)

\(n_1 + n_2-2=18 + 13-2 = 29\)

\((n_1 - 1)s_1^{2}=(18 - 1)\times40^{2}=17\times1600 = 27200\)

\((n_2 - 1)s_2^{2}=(13 - 1)\times25^{2}=12\times625 = 7500\)

\(s_p^{2}=\frac{27200 + 7500}{29}=\frac{34700}{29}\approx1196.5517\)

\(s_p=\sqrt{1196.5517}\approx34.591\)

The standard error for pooled variance is \(s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}=34.591\sqrt{\frac{1}{18}+\frac{1}{13}}\)

\(\frac{1}{18}+\frac{1}{13}=\frac{13 + 18}{234}=\frac{31}{234}\approx0.1325\)

\(\sqrt{0.1325}\approx0.364\)

\(s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}\approx34.591\times0.364\approx12.59\)

Then \(t=\frac{15}{12.59}\approx1.191\), which is not matching. Wait, maybe the original problem assumes unequal variances and my first calculation of the denominator was wrong. Let's recalculate the denominator:

\(\frac{40^2}{18}=\frac{1600}{18}\approx88.8889\)

\(\frac{25^2}{13}=\frac{625}{13}\approx48.0769\)

Sum: \(88.8889+48.0769 = 136.9658\)

Square root of 136.9658: \(\sqrt{136.9658}\approx11.703\)

Then \(t=\frac{15}{11.703}\approx1.282\), which is option D. Wait, but let's check with the correct formula for the two - sample t - test (unequal variances, Welch - Satterthwaite approximation for degrees of freedom, but the question is just about calculating the test statistic, not the degrees of freedom).

Wait, the numerator is \(\bar{x}_1-\bar{x}_2=605 - 590 = 15\)

Denominator: \(\sqrt{\frac{40^{2}}{18}+\frac{25^{2}}{13}}=\sqrt{\frac{1600}{18}+\frac{625}{13}}\approx\sqrt{88.8889 + 48.0769}=\sqrt{136.9658}\approx11.703\)

\(t=\frac{15}{11.703}\approx1.282\), which is option D.

Answer:

D. 1.282