Question

Suppose 40% of DC area adults have traveled outside of the United States. Nardole wants to know if his customers are atypical in this respect. He surveys 40 customers and finds 60% have traveled outside of the U.S. Is this result a statistically significant difference?

Answer 1

Answer:

We conclude that % of DC area adults who have traveled outside of the United States is different from 40%.

Step-by-step explanation:

We are given that 40% of DC area adults have traveled outside of the United States. Nardole wants to know if his customers are typical in this respect. He surveys 40 customers and finds 60% have traveled outside of the U.S.

We have to test is this result a statistically significant difference.

Let p = % of DC area adults who have traveled outside of the United States

SO, Null Hypothesis, $H_0$ : p = 40%  {means that 40% of DC area adults have traveled outside of the United States}

Alternate Hypothesis, $H_a$ : p $\neq$ 40%  {means that % of DC area adults who have traveled outside of the United States is different from 40%}

The test statistics that will be used here is One-sample z proportion statistics;

                  T.S. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1- \hat p)}{n} } }$  ~ N(0,1)

where, $\hat p$ = % of customers who have traveled outside of the United States

                  in a survey of 40 customers = 60%

          n  = sample of customers = 40

So, test statistics = $\frac{0.60-0.40}{\sqrt{\frac{0.60(1-0.60)}{40} } }$

                             = 2.582

Since in the question we are not given with the significance level so we assume it to be 5%. So, at 0.05 level of significance, the z table gives critical value of 1.96 for two-tailed test. Since our test statistics is more than the critical value of z so we have sufficient evidence to reject null hypothesis as it will fall in the rejection region.

Therefore, we conclude that % of DC area adults who have traveled outside of the United States is different from 40%.