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. (all hypothesis testing problems should include the null and alternative hypotheses and report the p-value of the data.)
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 the 10% level of significance?
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 no prior bounds are placed on the population proportion?

Explanation / Answer

PART A.

Given that,

possibile chances (x)=183

sample size(n)=300

success rate ( p )= x/n = 0.61

success probability,( po )=0.5

failure probability,( qo) = 0.5

null, Ho:p<0.5  

alternate, H1: p>0.5

level of significance, = 0.01

from standard normal table,right tailed z /2 =2.33

since our test is right-tailed

reject Ho, if zo > 2.33

we use test statistic z proportion = p-po/sqrt(poqo/n)

zo=0.61-0.5/(sqrt(0.25)/300)

zo =3.8105

| zo | =3.8105

critical value

the value of |z | at los 0.01% is 2.33

we got |zo| =3.811 & | z | =2.33

make decision

hence value of | zo | > | z | and here we reject Ho

p-value: right tail - Ha : ( p > 3.81051 ) = 0.00007

hence value of p0.01 > 0.00007,here we reject Ho

ANSWERS

---------------

null, Ho:p<0.5

alternate, H1: p>0.5

test statistic: 3.8105

critical value: 2.33

decision: reject Ho

p-value: 0.00007

majority of the employees are satisfied

PART C.

Given that,

possibile chances (x)=183

sample size(n)=300

success rate ( p )= x/n = 0.61

success probability,( po )=0.6

failure probability,( qo) = 0.4

null, Ho:p=0.6  

alternate, H1: p>0.6

level of significance, = 0.1

from standard normal table,right tailed z /2 =1.28

since our test is right-tailed

reject Ho, if zo > 1.28

we use test statistic z proportion = p-po/sqrt(poqo/n)

zo=0.61-0.6/(sqrt(0.24)/300)

zo =0.3536

| zo | =0.3536

critical value

the value of |z | at los 0.1% is 1.28

we got |zo| =0.354 & | z | =1.28

make decision

hence value of |zo | < | z | and here we do not reject Ho

p-value: right tail - Ha : ( p > 0.35355 ) = 0.36184

hence value of p0.1 < 0.36184,here we do not reject Ho

ANSWERS

---------------

null, Ho:p=0.6

alternate, H1: p>0.6

test statistic: 0.3536

critical value: 1.28

decision: do not reject Ho

p-value: 0.36184

does not supports the claim

PART D.

DIRECT METHOD

given that,

possibile chances (x)=183

sample size(n)=300

success rate ( p )= x/n = 0.61

CI = confidence interval

confidence interval = [ 0.61 ± 1.96 * Sqrt ( (0.61*0.39) /300) ) ]

= [0.61 - 1.96 * Sqrt ( (0.61*0.39) /300) , 0.61 + 1.96 * Sqrt ( (0.61*0.39) /300) ]

= [0.5548 , 0.6652]

PART E.

Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)

Z a/2 at 0.1 is = 1.645

Sample Proportion = 0.5

ME = 0.02

n = ( 1.645 / 0.02 )^2 * 0.5*0.5

= 1691.2656 ~ 1692

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