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 Carolina Statistical Abstract, sponsored by the South Carolina Budget and Control Board.

Part (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 \(75. Assume a population standard deviation of \)22.40.

Part (b): Obtain a normal probability plot, boxplot, histogram and stem-and-leaf diagram of the data.

Part (c): Remove the outliers (if any) from the data and then repeat part (a).

Part (d): Comment on the advisability of using thez-test here.

The null hypothesis is given below,

$H_0:\mu =\$75$

The data does not provide sufficient evidence to conclude that the mean daily charge for hotels and motels operating in South Carolina is less than $75.

The alternative hypothesis is given below,

$H_a:\mu >\$75$

The data provide sufficient evidence to conclude that the mean daily charge for hotels and motels operating in South Carolina is less than $75.

On computing the value of the test statistics,

If $P\le a$, then reject the null hypothesis.

Here, the P-value is 0.165 which is greater than the level of significance, that is $P\left(=0.165\right)>\alpha \left(=0.05\right)$.

Therefore, the null hypothesis is not rejected at 5% level.

Thus, it can be concluded that the results are not statistically significant at 5% level of significance.

On interpreting, we can say that the data does not provide sufficient evidence to conclude the mean daily charge for hotels and motels operating in South Carolina is less than$75.

On constructing a normal probability plot,

From the probability plot, the observations are closer to straight line with one outlier.

On constructing a boxplot,

From the boxplot, it is clear that the distribution of charge is right skewed with one outlier.

On constructing a histogram,

From the histogram, it is clear that the distribution of charge is right skewed.

On constructing a stem-and leaf diagram,

Steam-and left of CHARGE $N=15$

Leaf Unit$=1$

$HI130$

From the stem-and leaf diagram, it is clear that the distribution of charge is right skewed with one outlier.

The null hypothesis is given below,

$H_0:\mu =\$75$

The data does not provide sufficient evidence to conclude that the mean daily charge for hotels and motels operating in South Carolina is less than $75.

The alternative hypothesis is given below,

$H_a:\mu >\$75$

The data provide sufficient evidence to conclude that the mean daily charge for hotels and motels operating in South Carolina is less than $75.

On computing the value of the test statistics,

Therefore, the value of the test statistics is -1.67 and the P-value is 0.048.

If $P\le a$, then reject the null hypothesis.

Here, the P-value is 0.048 which is less than the level of significance, that is $P\left(=0.048\right)\le \alpha \left(=0.05\right)$.

Therefore, the null hypothesis is rejected at 5% level.

Thus, it can be concluded that the results are statistically significant at 5% level of significance.

On interpreting, we can say that the data provide sufficient evidence to conclude the mean daily charge for hotels and motels operating in South Carolina is less than $75.