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}=1.56 ) (round to two decimal places as needed.)
find the p - value.
p - value ( = 0.119 ) (round to three decimal places as needed)

Explanation:

Step1: Determine the hypotheses

The claim is that \(\mu_1>\mu_2\). The null hypothesis \(H_0\) is the statement of no difference or a non - favorable difference. The alternative hypothesis \(H_1\) is the claim we are testing. So \(H_0:\mu_1\leq\mu_2\) and \(H_1:\mu_1 > \mu_2\) (Option D).

Step2: Calculate the pooled variance

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

Step3: Calculate the test statistic

The formula for the \(t\) - test statistic for two independent samples with equal variances is \(t_{STAT}=\frac{\bar{X}_1-\bar{X}_2}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\)
\(\bar{X}_1 = 46\), \(\bar{X}_2=39\), \(S_p=\sqrt{53.5}\approx7.31\)
\(t_{STAT}=\frac{46 - 39}{7.31\sqrt{\frac{1}{8}+\frac{1}{4}}}=\frac{7}{7.31\sqrt{\frac{1 + 2}{8}}}=\frac{7}{7.31\sqrt{\frac{3}{8}}}\)
\(\sqrt{\frac{3}{8}}\approx0.612\), \(7.31\times0.612\approx4.47\)
\(t_{STAT}=\frac{7}{4.47}\approx1.56\)

Step4: Calculate the degrees of freedom

\(df=n_1 + n_2-2=8 + 4-2 = 10\)

Step5: Calculate the p - value

Using a \(t\) - distribution table or software (e.g., Excel: \(=T.DIST.RT(1.56,10)\)), the p - value is approximately \(0.119\)

Answer:

The hypotheses are \(H_0:\mu_1\leq\mu_2\) and \(H_1:\mu_1 > \mu_2\) (Option D). The test statistic \(t_{STAT}\approx1.56\) and the p - value is approximately \(0.119\)