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According to a recent report. 46% of college student internships are unpaid. A recent survey of 60 college interns at a local
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Answer #1

Given information:

n = 60, and a=29 where,

n= sample size

Ộ = 0.483333

where \mathbf{\hat{p} } = observed proportion

1) Problem statement

A report states that 46% of the college student internships are unpaid. To test this claim a random sample of 60 college interns is obtained. of these 29 interns had unpaid internships. is this claim accurate? use alpha= 0.01

solutions

1)Define the null and alternative hypothesis,

let x be the population proportion

Ho: x=0.46

H : x +0.46

2)What is the test statistics?

ZSTAT = P-Po po 1-po)

Po = 0.48333

p=0.46

n = 60

0.48333 - 0.46 ZSTAT = = 0.36258895 0.36 0.46(1-0.46) 60

3)Critical value

29 = 20.01 = 20.005

Here we have 0.005 in each tail. Looking up 1 - 0.005 in our z-table, we find a critical value of 2.58. Thus, our decision rule for this two-tailed test is: If Z is less than -2.58, or greater than 2.58, reject the null hypothesis. calculate Test Statistic:

  • using standard normal table

\mathbf{ Z_{0.005}=2.58}

  • Decision-based on critical value,
  • here we can see our ZStat value is less than 2.58 so we are going to accept the null hypothesis.

4) find the p-value

\mathbf{p(\hat{p}>0.4833|p=0.46)}=\mathbf{p(z>0.36)}

\mathbf{p(z>0.36)}\mathbf{=1-P(z<0.36)=1-0.6406=0.3594}

  • criteria to reject null hypothesis using p-value,

if p - value <a we reject the null hypothesis, i.e if

p - value < 0.01 we reject the null hypothesis otherwise we accept it.

  • Decision-based on P-value,

Here our p-value i.e 0.3594 > 0.01 we accept the null hypothesis.

5)Conclusion:

The proportion of college interns that had unpaid internships are not different from 0.46 i.e 46% of the college students internships are unpaid

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