91Ó°ÊÓ

$$\begin{gathered} {\stackrel{¯}{x}}_1=16177 \\ {\stackrel{¯}{x}}_2=16569 \\ {s}_1=7520 \\ {s}_2=9108 \\ {n}_1=56 \\ {n}_2=56 \end{gathered}$$

Determine the hypothesis:

$H_0:{\mathrm{μ}}_1={\mathrm{μ}}_2$

$H_a:{\mathrm{μ}}_1≠{\mathrm{μ}}_2$

Determine the test statistic:

localid="1650518880891" role="math" $$\begin{gathered} t=\frac{{\stackrel{¯}{x}}_1-{\stackrel{¯}{x}}_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}} \\ =\frac{16177-16569}{\sqrt{\frac{{7520}^2}{56}+\frac{{9108}^2}{56}}} \\ ≈-0.248 \end{gathered}$$

Determine the degrees of freedom:

localid="1650518751831" $$\begin{gathered} df=\min \left(n_1-1,n_2-1\right) \\ =\min (56-1,56-1) \\ =55>50 \end{gathered}$$

The P-value is the probability of obtaining the value of the test statistic, or a value more extreme. The P-value is the number (or interval) in the column title of Table B containing the t-value in the row$df=50$:

$P>2×0.25=0.50$

If the P-value is less than or equal to the significance level, then the null hypothesis is rejected:

$P>0.05⇒\text{Fail to reject}{H}_0$

There is not sufficient evidence to support the claim of a difference.

$$\begin{gathered} {\stackrel{¯}{x}}_1=16177 \\ {\stackrel{¯}{x}}_2=16569 \\ {s}_1=7520 \\ {s}_2=9108 \\ {n}_1=56 \\ {n}_2=56 \end{gathered}$$

Determine the hypothesis:

$H_0:{\mathrm{μ}}_1={\mathrm{μ}}_2$

$H_a:{\mathrm{μ}}_1≠{\mathrm{μ}}_2$

Determine the test statistic:

localid="1650518816454" $$\begin{gathered} t=\frac{{\stackrel{¯}{x}}_1-{\stackrel{¯}{x}}_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}} \\ =\frac{16177-16569}{\sqrt{\frac{{7520}^2}{56}+\frac{{9108}^2}{56}}} \\ ≈-0.248 \end{gathered}$$

Determine the degrees of freedom:

localid="1650518837853" $$\begin{gathered} df=\min \left(n_1-1,n_2-1\right) \\ =\min (56-1,56-1) \\ =55>50 \end{gathered}$$

The P-value is the probability of obtaining the value of the test statistic, or a value more extreme. The P-value is the number (or interval) in the column title of Table B containing the t-value in the row $df=20$:

localid="1650518859387" $$\begin{gathered} P>2×0.25=0.50 \\ =50\% \end{gathered}$$

This means that the probability of obtaining a sample with a mean difference of $16177-16569=-392$or more extreme is higher than $50\%$, if the population means are equal.