Question

assume that you have a sample of ( n_1 = 8 ), with the sample mean ( overline{x}_1 = 46 ), and a sample standard deviation of ( s_1 = 7 ), and you have an independent sample of ( n_2 = 4 ) from another population with a sample mean of ( overline{x}_2 = 39 ) and the sample standard deviation ( s_2 = 8 ). assuming the population variances are equal, at the 0.01 level of significance, is there evidence that ( mu_1>mu_2 )?
determine the hypotheses. choose the correct answer below.
a. ( h_0:mu_1=mu_2 )
( h_1:mu_1
eqmu_2 )
b. ( h_0:mu_1
eqmu_2 )
( h_1:mu_1=mu_2 )
c. ( h_0:mu_1>mu_2 )
( h_1:mu_1leqmu_2 )
d. ( h_0:mu_1leqmu_2 )
( h_1:mu_1>mu_2 )
find the test statistic.
( t_{stat}=square ) (round to two decimal places as needed)

Explanation:

Step1: Calculate the pooled variance

The formula for pooled variance \(S_p^2=\frac{(n_1 - 1)S_1^2+(n_2 - 1)S_2^2}{n_1 + n_2-2}\)
Substitute \(n_1 = 8\), \(S_1 = 7\), \(n_2=4\), \(S_2 = 8\)
\(S_p^2=\frac{(8 - 1)\times7^2+(4 - 1)\times8^2}{8 + 4-2}=\frac{7\times49+3\times64}{10}=\frac{343+192}{10}=\frac{535}{10}=53.5\)

Step2: Calculate the test statistic

The formula for the \(t\) - statistic in a two - sample \(t\) - test (equal variances) is \(t_{STAT}=\frac{\bar{X}_1-\bar{X}_2}{\sqrt{S_p^2(\frac{1}{n_1}+\frac{1}{n_2})}}\)
Substitute \(\bar{X}_1 = 46\), \(\bar{X}_2 = 39\), \(S_p^2=53.5\), \(n_1 = 8\), \(n_2 = 4\)
\(\sqrt{S_p^2(\frac{1}{n_1}+\frac{1}{n_2})}=\sqrt{53.5\times(\frac{1}{8}+\frac{1}{4})}=\sqrt{53.5\times(\frac{1 + 2}{8})}=\sqrt{53.5\times\frac{3}{8}}=\sqrt{\frac{160.5}{8}}\approx\sqrt{20.0625}=4.48\)
\(t_{STAT}=\frac{46 - 39}{4.48}=\frac{7}{4.48}\approx1.56\)

Answer:

\(t_{STAT}\approx1.56\)