Statistics Question Suppose that 88% of OSU college students were employed last

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Statistics Question

Suppose that 88% of OSU college students were employed last summer. A researcher interviews an SRS of 400 college students at OSU. In the SRS, 341 college students were found to be employed last summer. What is the sample proportion p^cap of college students at OSU who worked last summer? What is the approximate distribution of the proportion p^cap of college students at OSU who worked last summer? Using the 68 - 95 - 99.7 rule, if you had drawn an SRS of 400 college students from OSU, between what two percents would you expect p to fall about 95% of the time? Explain what is wrong in each of the following statements. For large sample size n, the distribution of observed values will he approximately normal. The 68-95-99.7 rule says that X^bar should be within mu plusorminus 2 sigma about 95% of the time. The central limit theorem states that for large n, mu is approximately normal.

Explanation / Answer

Sol:

2a:

sample proportion=p^=x/n=341/400=0.8525

Solution2b:

Approximate distribution is normal distribution as sample size =n>30 large sample

here n=400

Solutionc:

mean=p^

std deviation=sqrt[p^(1-p^)/n]

mean-2stddev

p^-2sqrt[p^(1-p^)/n]<p<p^+2sqrt[p^(1-p^)/n]

0.8525-2sqrt[0.8525(1-0.8525)/400]<p<0.8525+2sqrt[0.8525(1-0.8525)/400]

0.8170 <p<0.8702

we expect p^ to fall between 0.8170 and 0.8702 95% of the time.

Solution3:

The central limit theorem states that the sampling distribution of the mean of any independent, random variable will be normal or nearly normal, if the sample size is large enough.

Answser A is wrong

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