Answer :
Answer:
The answer is "between 0.10 and 0.05".
Step-by-step explanation:
For sample 1:
[tex]n_1 = 25\\\\\bar{x_1} = 90.00\\\\ s_1 = 9\\\\[/tex]
For sample 2:
[tex]n_2 = 25\\\\ \bar{x_2} = 95.08\\\\ S_2 = 12\\\\[/tex]
Calculating the null and alternative hypotheses:
[tex]H_O : \mu_1 = \mu_2 \\\\ H_Q : \mu_1 <\mu_2[/tex]
Calculating the test statistic:
[tex]Z= \frac{90-95.08}{ \sqrt{\frac{9^2}{25}+\frac{12^2}{25}}} \\\\ =\frac{-5.08}{ \sqrt{\frac{81}{25}+\frac{144}{25}}}\\\\=\frac{-5.08}{ \sqrt{\frac{81+144}{25}}}\\\\=\frac{-5.08}{ \sqrt{\frac{225}{25}}}\\\\=\frac{-5.08}{ \frac{15}{5}}\\\\=\frac{-5.08}{3}\\\\= -1.693[/tex]
Calculating the conservative degrees of freedom:
[tex]DF = min (n_1 -1, n_2 - 1) = min(25-1, 25-1) = 24[/tex]
by using Excel the p-value for left tailed test and for test statistic will be [tex]-1.693[/tex] with [tex]DF = 24[/tex].
[tex]\to p-value = TDIST(1.693, 24, 1) = 0.0517\ \ \ \ i.e, \bold{0.05 < p-value < 0.10}[/tex]