Question

A nationwide survey of working adults indicates that out of 100 adults, 50 of them are satisfied with their jobs. The president of a large company believes that more than this number of employees at his company are satisfies with their jobs. To test his belief, he surveys a random sample of 100 employees, and 59 of them report that they are satisfied with their jobs. Do you support president’s statement about the employees satisfaction? (Use α = 0.05 level of significance.)

(a) H0 :

(b) Ha :

(c) Test Statistic:

(d) Decision:

(e) Conclusion:

(f) Compute a 95% lower bound for population proportion p. Interpret your result.

Answer 1

Given:

A nationwide survey of working adults indicate that out of 100 adults 50 of them satisfied with their job

That is P=0.5

Here, X is Number of adults are satisfied with their job

Hence X follows binomial Distribution.

Given random samples of 100 employees and 59 reported that they are satisfied with their job.

Hence P_hat=X/n

Here n=100 and X=59

Hence P_hat=59/100=0.59

Now to find the E(P_hat)=0.5

V(P_hat)=P(1-P)/n

V(P_hat)=0.0025

Hence P_hat foll0ws Normal with mean=0.5 and variance=0.000025

Hence Z=(P_hat-P)/sqrt((P(1-P)/n)

Now

(a) H0:

H0: P=0.5

That is the Proportion of satisfied adults with their job is 0.5

(b) Ha:

Ha: P>0.5

That is the Proportion of satisfied adults with their job is greater than 0.5

(c) Test Statistic

In this type of situation. Test of equality of population proportion test is applicable

Hence Tets Statistics is

Z=(P_hat-P)/sqrt((P(1-P)/n)

Hence the value of Test statistics is 1.8

P	0.5
1-P	0.5
n	100
P_hat	0.59
Tets statistics	1.8

(d) Decision:

If Z>=Z(1-a) that is If Z_calculated is greater than Z tabulated then Reject H0 otherwise fail to reject H0

To find Z_tabulated

Here alpha=0.05 (Given)

Hence 1-alpha=1-0.05=0.95 see this value in the table (Normal) and take the corresponding value we get the Z tabulated value

and the required value is 1.64

Hence Z_calculate=1.8 is greater than Z_tabulated

Hence Reject H0. That is failing to reject H0

(e) Conclusion:-

That is the Proportion of satisfied adults with their job is greater than  0.5.

(f)

95% Lower bound for population proportion p is 0.492

Using Formula

Lower bound=P-hat-Z*sqrt((P*(1P))/n)

Hence Hence population proportion is larger of lower bound of CI of the population proportion.