9.90 Hotels and Motels. The daily charges, in dollars, for a sample of 15 hotels and motels operating in South Carolina are provided on the WeissStats site. The data were found in the report South Caroline Statistical Abstract, sponsored by the South Carolina Budget and Control Board.
a. Use the one-mean $z$-test to decide, at the $5\%$significance level, whether the data provide sufficient evidence to conclude that the mean daily charge for hotels and motels operating in South Carolina is less than $\backslash (75$. Assume a population standard deviation of $\backslash )22.40$.

b. Obtain a normal probability plot, boxplot, histogram, and stem-and-leaf diagram of the data.
c. Remove the outliers (if any) from the data and then repeat part (a).
d. Comment on the advisability of using the $z$-test here.

To use the one-mean $z$-test to decide, at the $5\%$significance level, whether the data provide sufficient evidence to conclude that the mean daily charge for hotels and motels operating in South Carolina is less than $\$75$.

The null hypothesis is indicated as:
$H_0:\mu =\$75$
The average daily fee for hostels and motels in South Carolina is less than $\$75$. So, the data does not support sufficient evidence.
The alternative hypothesis is indicated as:
$H_0:\mu <\$75$
The mean daily charge for hostel and motels in South Carolina is less than $\$75$. So, the data support sufficient evidence.
The significance level is $\alpha =0.05$.
MINITAB can be used to calculate the test statistic and P-value.
The output will be:

One sample Z: Charge

The test statistic is $-0.98$, and the P-value is $0.165$, according to the MINITAB output.
If $P\le \alpha$, the null hypothesis must be rejected.
The P-value is $0.165$, which is higher than the significance level.
$P(=0.165)>\alpha (=0.05)$
At the $5\%$ level, the null hypothesis is not rejected.
As a result, the data does not support the conclusion that the average daily fee for hostels and motels in South Carolina is less than $\$75$.

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

The Weiss Stats has supplied daily costs in dollars for a sample of 15 hotels and motels that are launching in South Carolina.
MINITAB is used to create the normal probability plot.
The output of MINITAB will be:

With one outlier, the observations appear to be closer to a straight line on the probability map.
MINITAB is used to create the boxplot.
The output of MINITAB will be:

The boxplot clearly shows that the charge distribution is right skewed with one outlier.
MINITAB is used to create the histogram.
The output of MINITAB will be:

The distribution of charge is right skewed with one outlier, as can be seen in the boxplot.
The output of MINITAB will be:
Stem-and-leaf Display: GAIN

The shape of the charge distribution is right skewed with one outlier, as shown in the stem and leaf diagram.
As a result, the shape of the charge distribution is right skewed with one outlier.

To remove the outliers from the data and then repeat part (a).

The null hypothesis is indicated as:
$H_0:\mu =\$75$
The data does not support the conclusion that the average daily fee for hostels and motels in South Carolina is less than $\$75$.
The alternative hypothesis is indicated as:
$H_0:\mu <\$75$
The data is sufficient to establish that the average daily fee for hostels and motels in South Carolina is less than $\$75$.
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: Charge

The test statistic has a value of $-1.67$and a $p$-value of $0.048$, according to the MINITAB output.
Reject the null hypothesis if $p\le \alpha$.
The P-value here is $0.048$, which is less than the significance level.
$P(=0.048)\le \alpha (=0.05)$
The null hypothesis is rejected at $5\%$ level.
As a result of the data, it is reasonable to determine that the average daily price for hostels and motels in South Carolina is less than $\$75$.

To comment on the advisability of using the $z$-test.

The Weiss Stats has supplied daily costs in dollars for a sample of $15$ hotels and motels that are launching in South Carolina.
The sample size for the original data appears to be $15$ based on the above mentioned results.
One outlier $(=130.17)$ emerges in the data from part (b).
After removing the outlier in part (c), the $z$-test is calculated, yielding the test statistic.
$-0.98\text{to}-1.67$
As a result, after eliminating the outlier charge from the original data, the $z$-test is calculated.