91Ó°ÊÓ

To obtain a normal probability plot, boxplot, histogram, and stemand-leaf diagram of the data.

On the Weiss Stats site, the local monthly costs in dollars for a random sample of $75$ call phone users occurs from the previous year.
MINITAB is used to create the normal probability plot.
The output of the MINITAB will be:

The observations are closer to a straight line on the probability plot, with two outliers.
MINITAB is used to create the boxplot.
The output of MINITAB will be:

The boxplot shows that the bill distribution is right skewed, with two outliers.
MINITAB is used to create the histogram.
The output of MINITAB will be:

The histogram clearly shows that the bill distribution is right skewed, with two outliers.
Using MINITAB, create the stem-and-leaf diagram.
The output of MINITAB will be:
Stem-and-leaf Display: BILL

The shape of the bill distribution is right skewed with two outliers, as shown in the stem and leaf diagram.
As a result, the bill distribution is right skewed with two outliers.

Let, the population standard deviation of last year's local monthly bills for cell phone users is $\$25$.

The null hypothesis is follows as:
$H_0:\mathrm{μ}=\$48.16$
The data does not support the conclusion that local monthly bills for cell phone users reduced last year from the $2009$ mean of $\$48.16$.
The alternative hypothesis is follows as:
$H_0:\mathrm{μ}<\$48.16$
The data are adequate to determine that local monthly bills for cell phone users declined last year, compared to the $2009$ mean of $\$48.16$.
The significance level is $\mathrm{Î\pm }=0.05$.
MINITAB can be used to calculate the test statistic and P-value.
The output of MINITAB will be:
One sample Z: BILL

The test statistic value is $-0.35$, and the $p$value is $0.365$, according to the MINITAB output.
If $p≤\mathrm{Î\pm }$, the null hypothesis must be rejected.
The $p$-value is $0.365$, which is higher than the significance level.
$P(=0.365)>\mathrm{Î\pm }(=0.05)$
At the $5\%$ level, the null hypothesis is not rejected.
As a result, the data does not support the conclusion that last year's local monthly bills for cell phone customers were lower than the $2009$ mean of $\$48.16$.

To remove the two outliers from the data and repeat parts (a) and (b).

On the Weiss Stats site, the local monthly costs in dollars for a random sample of $75$ call phone users occurs from the previous year.
MINITAB is used to create the normal probability plot.
The output of the MINITAB will be:

The observations are closer to a straight line on the probability plot, with two outliers.
MINITAB is used to create the boxplot.
The output of MINITAB will be:

The histogram clearly shows that the bill distribution is right skewed, with two outliers.
Using MINITAB, create the stem-and-leaf diagram.
The output of MINITAB will be:
Stem-and-leaf Display: BILL

The shape of the charge distribution is right skewed with one outlier, as shown in the stem and leaf diagram.
As a result, the charge distribution has a right-skewed form with one outlier.
The null hypothesis is indicated as follows:
$H_0:\mathrm{μ}=\$48.16$
The data does not support the conclusion that local monthly bills for cell phone users reduced last year from the $2009$mean of $\$48.16$.
The alternative hypothesis is indicated as follows:
$H_0:\mathrm{μ}<\$48.16$
The data are adequate to determine that local monthly bills for cell phone users declined last year, compared to the $2009$mean of $\$48.16$.
The significance level is $\mathrm{Î\pm }=0.05$.
MINITAB can be used to calculate the test statistic and P-value.

The output of MINITAB will be:
One sample Z: BILL

The test statistic is $-0.97$, and the P-value is $0.167$, according to the MINITAB output.
If $p≤\mathrm{Î\pm }$, the null hypothesis must be rejected.
The $p$-value is $0.167$, which is higher than the significance level.
$P(=0.167)>\mathrm{Î\pm }(=0.05)$
At the $5\%$ level, the null hypothesis is not rejected.
As a result, the data does not support the conclusion that last year's local monthly bills for cell phone customers were lower than the $2009$ mean of $\$48.16$.

To state the conclusions regarding the hypothesis test.

On the Weiss Stats site, the local monthly costs in dollars for a random sample of $75$ call phone users occurs from the previous year.
The sample size for the original data is $75$, as shown by the preceding results.
There are two outliers in the data from part (a).
After removing the outlier in part (c), the z-test is calculated, yielding the test statistic.
$-0.35\text{to}-0.97$
As a result, the $z$-test is calculated after the outlier is removed, and it does not affect the original data.