Testing a hypothesys about a survey? A survey of 61,647 people including questions about office relationships. Of the respondents, 26% reported that bosses scream at employees. Use a 05 significance level to test the claim that 1/4 of the people say that bosses scream at employees. Please keep things as simple as possible. My work: H_0:p=25 H_1:p>25 P is given as 26 Using the formula z=P(hat)\(sqrt(pq\n)) with q=75 , I get a z-score of around 5.73. However, this is an extreme value but 26 percent is very close to percent.25, so I was expecting a z score of almost zero. What did I do wrong?

cochetezgh

cochetezgh

Answered question

2022-09-08

Testing a hypothesys about a survey?
A survey of 61,647 people including questions about office relationships. Of the respondents, 26% reported that bosses scream at employees. Use a 05 significance level to test the claim that 1/4 of the people say that bosses scream at employees.
Please keep things as simple as possible.
My work: H 0 : p = .25
H 1 : p > .25
P is given as 26
Using the formula z = P ^ p q n with q=75, I get a z-score of around 5.73. However, this is an extreme value but percent is very close to percent., so I was expecting a z score of almost zero. What did I do wrong?

Answer & Explanation

Annie Wells

Annie Wells

Beginner2022-09-09Added 17 answers

Under the null hypothesis, p = 0.25. The test you are using (appropriately, I might add), relies on modeling the sampling distribution of P ^ as N ( μ = p , σ = p q n ) .
The test you are referring to is not correctly stated in your post, it should be z = P ^ 0.25 p q n ,which does in fact give you 5.73. As Henry has noted, this results from your very large sample size, which will have a very high power. Thus, it can detect even very small deviations from 25%.
The 95% confidence interval for your true p would be approx. 0.26 ± 1.96 ( .002 ) 0.26 ± 003. So you can be quite confident that p=0.26 (to two significant figures.) So yeah, its different from 0.25 but not by very much.

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