The Human Resources Department of a large corporation would like to determine if

Question

The Human Resources Department of a large corporation would like to determine if a majority of its employees were satisfied with their treatment by the corporation's health care provider. A random sample of 300 employees was selected, and 183 indicated that they were satisfied with their treatment. a) Does this data show at the 1% level of significance that a majority of all employees are satisfied? b) What is the probability that this test would reject the null hypothesis if the actual proportion of satisfied employees is 0.55? c Suppose the Corporation's president requires that more than 60% of employees should be satisfied. Does this data support that requirement at e 0% ee lofs n i ance? d) Construct a 95% confidence interval for the proportion of all employees who are satisfied. e) What sample size would be required to estimate this proportion to within 2% with 90% confidence if n prior bounds are ace on the po ula on proportion

Explanation / Answer

a)
pcap = 0.61

Below are the null and alternate hypo
H0: p = 0.5
H1: p > 0.5

Test statistics z = (0.61 - 0.5)/sqrt(0.5*0.5/300) = 3.8105

p-value = 0.000069

As p-value is less than significance level of 0.01, we reject the null hypothesis.
This means the data shows that a majority of all employees are satisfied.

b)

c)
pcap = 0.61

Below are the null and alternate hypo
H0: p = 0.6
H1: p > 0.6

Test statistics z = (0.61 - 0.6)/sqrt(0.6*0.4/300) = 0.3536

p-value = 0.3618

As p-value is greater than significance level of 0.01, we fail to reject the null hypothesis.
This means the data does not show that a more than 60% of employees are satisfied.

d)

e)

p0 (hypothesised proportion) 0.5 SE = sqrt(p*(1-p)/n) 0.02886751 n 300 alpha 0.01 sample/true proportion 0.55 Std. Error. SE 0.0289 Zcritical 2.326 Xcritical 0.57 Beta or type II error is the probability of fail to reject the null hypothesis P(p<0.57) 0.2762 Hence type II error probability is 0.2762

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