Given: y || z Prove: m<5 + m<2 + m<6 + 180
1 answer:
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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.
<em>Let p = % of DC area adults who have traveled outside of the United States</em>
SO, Null Hypothesis, : p = 40% {means that 40% of DC area adults have traveled outside of the United States}
Alternate Hypothesis, : p
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 <u>One-sample z proportion statistics</u>;
T.S. = ~ N(0,1)
where, = % of customers who have traveled outside of the United States
in a survey of 40 customers = 60%
n = sample of customers = 40
So, <u>test statistics</u> =
= 2.582
<em>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.</em>
Therefore, we conclude that % of DC area adults who have traveled outside of the United States is different from 40%.