91Ó°ÊÓ

$$\begin{gathered} {\stackrel{¯}{x}}_1=6.37 \\ {\stackrel{¯}{x}}_2=5.91 \\ {n}_1=92 \\ {n}_2=86 \\ {s}_1=0.60 \\ {s}_2=0.93 \end{gathered}$$

The given claim is that there is a disparity in the means.

Now we must determine the most appropriate hypotheses for a significance test.

As a result, either the null hypothesis or the alternative hypothesis is the claim. According to the null hypothesis, the population proportions are equal. If the claim is the null hypothesis, the alternative hypothesis is the polar opposite of the null hypothesis.

Therefore, the appropriate hypotheses for this are:

$$\begin{gathered} H_0:{\mathrm{μ}}_1={\mathrm{μ}}_2 \\ {H}_a:{\mathrm{μ}}_{1}notequalto{\mathrm{μ}}_2 \end{gathered}$$

Where we have,

${\mathrm{μ}}_1=$the true mean of all Angle bank customers' reliability ratings.

${\mathrm{μ}}_2=$is the true mean of all Hispanic bank customers' reliability ratings.

We then know that the sample means $6.37$and $5.91$difference which agrees with the alternative hypothesis $H_a:{\mathrm{μ}}_1≠{\mathrm{μ}}_2$that the means are different and thus the sample results give evidence for the alternative hypothesis.

$$\begin{gathered} {\stackrel{¯}{x}}_1=6.37 \\ {\stackrel{¯}{x}}_2=5.91 \\ {n}_1=92 \\ {n}_2=86 \\ {s}_1=0.60 \\ {s}_2=0.93 \end{gathered}$$

The given claim is that there is a disparity in the means.

Now we must determine the most appropriate hypotheses for a significance test.

As a result, either the null hypothesis or the alternative hypothesis is the claim. According to the null hypothesis, the population proportions are equal. If the claim is the null hypothesis, the alternative hypothesis is the polar opposite of the null hypothesis.

Therefore, the appropriate hypotheses for this are:

$$\begin{gathered} H_0:{\mathrm{μ}}_1={\mathrm{μ}}_2 \\ {H}_a:{\mathrm{μ}}_{1}notequalto{\mathrm{μ}}_2 \end{gathered}$$

Where we have,

${\mathrm{μ}}_1=$the true mean of all Angle bank customers' reliability ratings.

${\mathrm{μ}}_2=$is the true mean of all Hispanic bank customers' reliability ratings.

Locate the following test statistics:

$$\begin{gathered} t=\frac{({\stackrel{¯}{x}}_1-{\stackrel{¯}{x}}_2)-({\mathrm{μ}}_1-{\mathrm{μ}}_2)}{\sqrt{{\displaystyle \frac{{s_1}^2}{n_1}}+{\displaystyle \frac{{s_2}^2}{n_2}}}} \\ =\frac{6.37-5.91-0}{\sqrt{{\displaystyle \frac{0.{60}^2}{92}}+{\displaystyle \frac{0.{93}^2}{86}}}} \\ =3.892 \end{gathered}$$

Now, the degree of freedom will be:

$$\begin{gathered} df=min(n_1-1,n_2-1) \\ =min(92-1,86-1) \\ =85 \end{gathered}$$

Hence, the student's T distribution table in the appendix does not contain the value of $df=85$so we will take the nearest value So the $P$-value will be:

$P<2(0.0005)=0.001$

On the other hand by using the calculator command: $2×tcdf(3.892,1\mathrm{E}99,85)$) which results in the $P$-values as $0.0002$
Therefore, the $P$-value is $P<0.001$or$P=0.0002$and the$t=3.892$

From parts (a) and (b), we have,

$$\begin{gathered} {\stackrel{¯}{x}}_1=6.37 \\ {\stackrel{¯}{x}}_2=5.91 \\ {n}_1=92 \\ {n}_2=86 \\ {s}_1=0.60 \\ {s}_2=0.93 \end{gathered}$$

For this, the following hypotheses are appropriate:

$$\begin{gathered} H_0:{\mathrm{μ}}_1={\mathrm{μ}}_2 \\ {H}_a:{\mathrm{μ}}_{1}notequalto{\mathrm{μ}}_2 \end{gathered}$$

And the $P$-value is $P<0.001$or $P=0.0002$and the $t=3.892$

And we know that the null hypothesis is rejected if the P-value is less than or equal to the significance level.

$P<0.05⇒\text{Reject}{H}_0$

Therefore, we conclude that the mean reliability ratings of all Angle and Hispanic bank customers differ significantly.