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?

The p value is a very low value and using any significance level for example [tex]\alpha=0.05, 0,1,0.15[/tex] always [tex]p_v<\alpha[/tex] 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.

Step-by-step explanation:

1) Data given and notation

[tex]X_{B}=178[/tex] represent the number of requested nonsmoking rooms before

[tex]X_{A}=320[/tex] represent the number of requested nonsmoking rooms after

[tex]n_{B}=378[/tex] sample of number of requested nonsmoking rooms before

[tex]n_{A}=516[/tex] sample of number of requested nonsmoking rooms after

[tex]p_{B}=\frac{178}{378}=0.471[/tex] represent the proportion of requested nonsmoking rooms before

[tex]p_{A}=\frac{320}{516}=0.620[/tex] represent the proportion of requested nonsmoking rooms after

z would represent the statistic (variable of interest)

[tex]p_v[/tex] 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:[tex]p_{A} \leq p_{B}[/tex]

Alternative hypothesis:[tex]p_{A} > p_{B}[/tex]

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

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

Where [tex]\hat p=\frac{X_{A}+X_{B}}{n_{A}+n_{B}}}=\frac{178+320}{378+516}=0.557[/tex]

3) Calculate the statistic

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

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

4) Statistical decision

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

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

[tex]p_v =P(Z>4.43)=4.71x10^{-6}[/tex]

So the p value is a very low value and using any significance level for example [tex]\alpha=0.05, 0,1,0.15[/tex] always [tex]p_v<\alpha[/tex] 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.