Question

find the standardized test statistic, t, to test the claim that $\mu_1 \
eq \mu_2$. two samples are r populations that are normal. the sample statistics are given below. assume that $\sigma_1^2 \
eq$
$n_1 = 11, n_2 = 18, \bar{x}_1 = 7.4, \bar{x}_2 = 7.8, s_1 = 0.76, s_2 = 0.51$

\\(\bigcirc\\) a. \\( -1.326
\bigcirc\\) b. \\( -2.123
\bigcirc\\) c. \\( -1.821
\bigcirc\\) d. \\( -1.546
\\)

Explanation:

Step1: Recall the formula for t - statistic (unequal variances)

The formula for the t - statistic when testing the claim that \(\mu_1
eq\mu_2\) with unequal variances (\(\sigma_1^2
eq\sigma_2^2\)) 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 the claim is \(\mu_1
eq\mu_2\), the null hypothesis is \(H_0:\mu_1 = \mu_2\), so \(\mu_1-\mu_2 = 0\). The formula simplifies to:

$$t=\frac{\bar{x}_1-\bar{x}_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}$$

Step2: Substitute the given values

We are given \(n_1 = 11\), \(n_2=18\), \(\bar{x}_1 = 7.4\), \(\bar{x}_2=7.8\), \(s_1 = 0.76\), \(s_2 = 0.51\)

First, calculate the numerator: \(\bar{x}_1-\bar{x}_2=7.4 - 7.8=- 0.4\)

Then, calculate the denominator:
\(\frac{s_1^2}{n_1}=\frac{(0.76)^2}{11}=\frac{0.5776}{11}\approx0.0525\)
\(\frac{s_2^2}{n_2}=\frac{(0.51)^2}{18}=\frac{0.2601}{18}\approx0.01445\)
\(\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}=\sqrt{0.0525 + 0.01445}=\sqrt{0.06695}\approx0.2587\)

Step3: Calculate the t - statistic

$$t=\frac{- 0.4}{0.2587}\approx - 1.546$$

Answer:

D. - 1.546