Respuesta :
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, [tex]H_0[/tex] : p = 40% {means that 40% of DC area adults have traveled outside of the United States}
Alternate Hypothesis, [tex]H_a[/tex] : p [tex]\neq[/tex] 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. = [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1- \hat p)}{n} } }[/tex] ~ N(0,1)
where, [tex]\hat p[/tex] = % 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 = [tex]\frac{0.60-0.40}{\sqrt{\frac{0.60(1-0.60)}{40} } }[/tex]
= 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%.
