91Ó°ÊÓ

$$\begin{gathered} {\stackrel{¯}{x}}_1=17.6 \\ {\stackrel{¯}{x}}_2=9.5 \\ {s}_1=6.34 \\ {s}_2=1.95 \\ {n}_1=6 \\ {n}_2=6 \end{gathered}$$

Determine the hypothesis:

$H_0:{\mathrm{μ}}_1={\mathrm{μ}}_2$

$H_a:{\mathrm{μ}}_1≠{\mathrm{μ}}_2$

Determine the test statistic:

localid="1650519030821" $$\begin{gathered} t=\frac{{\stackrel{¯}{x}}_1-{\stackrel{¯}{x}}_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}} \\ =\frac{17.6-9.5}{\sqrt{\frac{6.{34}^2}{6}+\frac{1.{95}^2}{6}}} \\ ≈2.291 \end{gathered}$$

Determine the degrees of freedom:

localid="1650519055597" $$\begin{gathered} df=\min \left(n_1-1,n_2-1\right) \\ =\min (6-1,6-1) \\ =5 \end{gathered}$$

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 $df=5$:

$0.02=2×0.01<P<2×0.02=0.04$

(In the output we note that the exact P-value is$0.0247$).

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

$P<0.05⇒\text{Reject}{H}_0$

There is sufficient evidence to support the claim of a difference.

$$\begin{gathered} {\stackrel{¯}{x}}_1=17.6 \\ {\stackrel{¯}{x}}_2=9.5 \\ {s}_1=6.34 \\ {s}_2=1.95 \\ {n}_1=6 \\ {n}_2=6 \end{gathered}$$

Determine the hypothesis:

$H_0:{\mathrm{μ}}_1={\mathrm{μ}}_2$

$H_a:{\mathrm{μ}}_1≠{\mathrm{μ}}_2$

Determine the test statistic:

localid="1650519139340" $$\begin{gathered} t=\frac{{\stackrel{¯}{x}}_1-{\stackrel{¯}{x}}_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}} \\ =\frac{17.6-9.5}{\sqrt{\frac{6.{34}^2}{6}+\frac{1.{95}^2}{6}}} \\ ≈2.29 \end{gathered}$$

Determine the degrees of freedom:

localid="1650519154640" $$\begin{gathered} df=\min \left(n_1-1,n_2-1\right) \\ =\min (6-1,6-1) \\ =5 \end{gathered}$$

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 $df=20$:

$0.02=2×0.01<P<2×0.02=0.04$

In the output, we note that the exact P-value is localid="1651483737374" $0.0247=2.47\%.$

This means that the probability of obtaining a sample with a mean difference of $17.6-9.5=8.1$or more extreme is equal to $2.47\%$, if the population means are equal.