Question

9.93 Cell Phones. The number of cell phone users has increased dramatically since $1987$. According to the Semi-annual Wireless Survey, published by the Cellular Telecommunications & Internet Association, the mean local monthly bill for cell phone users in the United States was $\backslash (48.16$in $2009$ . Last year's local monthly bills, in dollars, for a random sample of $75$ cell phone users are given on the WeissStats site. Use the technology of your choice to do the following.
a. Obtain a normal probability plot, boxplot, histogram, and stemand-leaf diagram of the data.
b. At the $5\%$significance level, do the data provide sufficient evidence to conclude that last year's mean local monthly bill for cell phone users decreased from the $2009$ mean of $\backslash )48.16?$Assume that the population standard deviation of last year's local monthly bills for cell phone users is $\$25$.
c. Remove the two outliers from the data and repeat parts (a) and (b).
d. State your conclusions regarding the hypothesis test.

Short answer

(a) The bill distribution is right skewed with two outliers.

(b) 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 $\$48.16$.

(c) With one outlier, the shape of the charge distribution is right skewed.
The data does not support the conclusion that local monthly bills for cell phone users reduced last year from the $2009$mean of $\$48.16$.
(d) The $z$-test is calculated after the outlier is removed, and it does not affect the original data.

Step-by-step solution

Part (a) Step 1: Given information

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

Part (a) Step 2: Explanation

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:

Part (a) Step 3: Explanation

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:

Part (a) Step 4: Explanation

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:

Part (a) Step 5: Explanation

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.

Part (b) Step 1: Given information

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

Part (b) Step 2: Explanation

The null hypothesis is follows as:
$H_0:\mu =\$48.16$
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:
$H_0:\mu <\$48.16$
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 $\alpha =0.05$.
MINITAB can be used to calculate the test statistic and P-value.
The output of MINITAB will be:
One sample Z: BILL

Part (b) Step 3: Explanation

The test statistic value is $-0.35$, and the $p$value is $0.365$, according to the MINITAB output.
If $p\le \alpha$, the null hypothesis must be rejected.
The $p$-value is $0.365$, which is higher than the significance level.
$P(=0.365)>\alpha (=0.05)$
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 $\$48.16$.

Part (c) Step 1: Given information

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

Part (c) Step 2: Explanation

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:

Part (c) Step 3: Explanation

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:

Part (c) Step 4: Explanation

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

Part (c) Step 5: Explanation

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:
$H_0:\mu =\$48.16$
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:
$H_0:\mu <\$48.16$
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 $\alpha =0.05$.
MINITAB can be used to calculate the test statistic and P-value.

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

Part (c) Step 6: Explanation

The test statistic is $-0.97$, and the P-value is $0.167$, according to the MINITAB output.
If $p\le \alpha$, the null hypothesis must be rejected.
The $p$-value is $0.167$, which is higher than the significance level.
$P(=0.167)>\alpha (=0.05)$
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 $\$48.16$.

Part (d) Step 1: Given information

To state the conclusions regarding the hypothesis test.

Part (d) Step 2: Explanation

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.