Question

A random sample of 378 hotel guests was taken one year ago, and it was found that 178 requested nonsmoking rooms. Recently, a random sample of 516 hotel guests showed that 320 requested nonsmoking rooms. Do these data indicate that the proportion of hotel guests requesting nonsmoking rooms has increased?

Answer 1

The p value is a very low value and using any significance level for example
`\alpha=0.05, 0,1,0.15` always
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the the proportion of proportion of hotel guests requesting nonsmoking rooms has increased.

Explanation:

1) Data given and notation

`X_(B)=178` represent the number of requested nonsmoking rooms before

`X_(A)=320` represent the number of requested nonsmoking rooms after

`n_(B)=378` sample of number of requested nonsmoking rooms before

`n_(A)=516` sample of number of requested nonsmoking rooms after

`p_(B)=(178)/(378)=0.471` represent the proportion of requested nonsmoking rooms before

`p_(A)=(320)/(516)=0.620` represent the proportion of requested nonsmoking rooms after

z would represent the statistic (variable of interest)

`p_v` represent the value for the test (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to check if the proportion of requested nonsmoking rooms after is higher then the proportion before , the system of hypothesis would be:

Null hypothesis:
`p_(A) \leq p_(B)`

Alternative hypothesis:
`p_(A) > p_(B)`

We need to apply a z test to compare proportions, and the statistic is given by:

`z=\frac{p_(A)-p_(B)}{\sqrt{\hat p (1-\hat p)((1)/(n_(A))+(1)/(n_(B)))}}` (1)

Where
`\hat p=(X_(A)+X_(B))/(n_(A)+n_(B))}=(178+320)/(378+516)=0.557`

3) Calculate the statistic

Replacing in formula (1) the values obtained we got this:

`z=\frac{0.620-0.471}{\sqrt{0.557(1-0.557)((1)/(516)+(1)/(378))}}=4.43`

4) Statistical decision

For this case we don't have a significance level provided
`\alpha`, but we can calculate the p value for this test.

Since is a one side test the p value would be:

`p_v =P(Z>4.43)=4.71x10^(-6)`

So the p value is a very low value and using any significance level for example
`\alpha=0.05, 0,1,0.15` always
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the the proportion of proportion of hotel guests requesting nonsmoking rooms has increased.