91影视

\(\begin{array}{l}{{\bar x}_1} = 2.7186\\{{\bar x}_2} = 2.8639\\{s_1} = 0.63342\\{s_2} = 0.49241\end{array}\)

\(\begin{array}{l}{n_1} = 125\\{n_2} = 88\\\alpha = 0.01\end{array}\)

We may use the \(z\)-test because the samples are large \(\left( {n > 30} \right)\). (instead of a t-test).

Assumption:

Either the null hypothesis or the alternative hypothesis is asserted. The null hypothesis and the alternative hypothesis are diametrically opposed. An equality must be included in the null hypothesis.

\(\begin{array}{l}{H_0}:{\mu _1} = {\mu _2}\\{H_a}:{\mu _1} \ne {\mu _2}\end{array}\)

Determine the value of the test statistic:

\(z = \frac{{{{\bar x}_1} - {{\bar x}_2}}}{{\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} }} = \frac{{2.7186 - 2.8639}}{{\sqrt {\frac{{0.6334{2^2}}}{{125}} + \frac{{0.4924{1^2}}}{{88}}} }} \approx - 1.88\)

If the null hypothesis is true, the \(P\)-value is the probability of getting a result more extreme or equal to the standardized test statistic \(z\). Using the normal probability table, determine the probability.

\(P = P(Z < - 1.88 or Z > 1.88) = 2P(Z < - 1.88) = 2(0.0301) = 0.0602\)

The null hypothesis is rejected if the P-value is less than the alpha significance level.

\(P > 0.01 \Rightarrow \)fail to reject \({H_0}\)

There is insufficient data to support the notion that part-time faculty have a different average course GPA than full-time faculty.

There is little evidence to support the notion that part-time faculty's genuine average course GPA differs from that of full-time professors.