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

Answer:

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.  

Step-by-step 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

$p_{B}=\frac{178}{378}=0.471$ represent the proportion of requested nonsmoking rooms before

$p_{A}=\frac{320}{516}=0.620$ 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:  

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

Where $\hat p=\frac{X_{A}+X_{B}}{n_{A}+n_{B}}}=\frac{178+320}{378+516}=0.557$  

3) Calculate the statistic

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

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

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.