Question

Of the 17.4 million overnight stays in Denver 477,000 guests chose Airbnb. This represents 2.74% of guests choose an Airbnb. In contrast, of the 8.420 billion overnight stays in the United States 33.9 million choose an Airbnb. This represents .40% of guests choose an Airbnb. Is there a statistical difference of Airbnb users in Denver compared to the United States?

Answer 1

The following information is provided: The sample size is N=17400000, the number of favorable cases is X=477000, and the sample proportion is

, and assuming the significance level is α=0.05

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho:

Ha:

This corresponds to a two-tailed test, for which a z-test for one population proportion needs to be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the critical value for a two-tailed test is zc​=1.96.

The rejection region for this two-tailed test is R={z:∣z∣>1.96}

(3) Test Statistics

The z-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that ∣z∣=1547.344>zc​=1.96, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0, and since p=0<0.05, it is concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population proportion p is different than p0​, at the α=0.05 significance level.

Graphically

Please let me know in comments in case anything is unclear. Will reply ASAP. Do upvote if satisfied!