Question

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.

Short answer

Part (a): 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.

Part (b): On constructing a normal probability plot, we get,

On constructing a boxplot,

On constructing a histogram,

On constructing a stem-and leaf diagram,

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

Part (d): From part (b) there is one outlier $\left(=130.17\right)$appears in the data.

From part (c), the z-test is calculate after removing the outlier, which provides the test statics 1.01 to 1.75.

Step-by-step solution

Part (a) Step 1. Given information.

Consider the given question,

A population standard deviation is $22.40.

Part (a) Step 2. State the null and alternative hypothesis.

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 -0.98 and the P-value is 0.165.

Part (a) Step 3. Interpret the result.

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.

Part (b) Step 1. Construct a normal probability plot, boxplot of the data.

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.

Part (b) Step 2. Construct a histogram and stem-and-leaf diagram of the data.

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.

Part (c) Step 1. State the null and alternative hypothesis and remove the outliers.

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.

Part (c) Step 2. Interpret the result.

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.

Part (d) Step 1. Comment on the advisability of using the z-test here.

Consider the results, it is clear that the sample size for the original data is 15.

From part (b) there is one outlier $\left(=130.17\right)$appears in the data.

From part (c), the z-testis calculate after removing the outlier, which provides the test statics -0.98 to -1.67.

Hence, the z-test is calculated after removing the outlier change the conclusion of the original data.