Question

A survey of 61,649 people included several questions about office relationships. Of the​ respondents, 27.2​% reported that bosses scream at employees. Use a 0.01 significance level to test the claim that more than 1/4 of people say that bosses scream at employees. How is the conclusion affected after learning that the survey is an online survey in which Internet users chose whether to​ respond? Identify the null​hypothesis, alternative​ hypothesis, test​ statistic, P-value, conclusion about the null​ hypothesis, and final conclusion that addresses the original claim. Use the​ P-value method. Use the normal distribution as an approximation of the binomial distribution.

Answer 1

Answer)

1/4 = 0.25

Null hypothesis Ho : P = 0.25

Alternate hypothesis Ha : P > 0.25

N = 61649

P = 0.272

As the sample size is extremely large, we can use standard normal z table, to conduct the test

Test statistics z = (observed P - Claimed P)/standard error

Standard error = √{claimed p*(1-claimed p)}/√n

Observed P = 0.272

Claimed P = 0.25

N = 61649

After substitution

Z = 12.6149363322

Z = 12.61

From z table, P(z>12.61) = 0

Therefore, P-Value is = 0

As the obtained P-Value is less than the given significance level of 0.01

We reject the null hypothesis Ho

And so we have enough evidence to support the claim that, P > 0.25 (1/4)