Question

A report claims that 46​% of​ full-time college students are employed while attending college. A recent survey of 60 ​full-time students at a state university found that 28 were employed. Use the​ five-step p-value approach to hypothesis testing and a 0.05 level of significance to determine whether the proportion of​ full-time students at this state university is different from the national norm of 0.46. find the t statistic find the p value find the critical value

Answer 1

Answer)

Null hypothesis Ho : P = 0.46

Alternate hypothesis Ha : P not equal to 0.46

N = 60

First we need to check the conditions of normality that is if n*p and n*(1-p) both are greater than 5 or not

N*p = 27.6

N*(1-p) = 32.4

Both the conditions are met so we can use standard normal z table to estimate the P-Value

Test statistics z = (oberved p - claimed p)/standard error

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

Observed P = 28/60

N = 60

Claimed P = 0.46

Z = 0.1

From z table, P(z>0.1) = 0.4602

As the test is two tailed

P-value = 2*0.4602 = 0.9204

As the obtained p-value is greater than 0.05 given significance

We fail to reject the null hypothesis Ho

We do not have enough evidence to support the claim that proportion is different from 0.46

Given significance level is 0.05

As the test is two tailed

We will divide it into two equal parts

0.025

From z table, P(z<-1.96) = P(z>1.96) = 0.025

So, critical values are -1.96 and 1.96

Rejection region is reject Ho if test statistics is less than -1.96 or greater than 1.96

As 0.1 is not less than -1.96 nor greater than 1.96

We fail to reject the null hypothesis