Question

Life after college. We are interested in estimating the proportion of graduates at a mid-sized university who found a job within one year of completing their undergraduate degree. Suppose we conduct a survey and find out that 348 of the 400 randomly sampled graduates found jobs. The graduating class under consideration included over 4500 students.(a) Describe the population parameter of interest. What is the value of the point estimate of this parameter?(b) Check if the conditions for constructing a confidence interval based on these data are met.(c) Calculate a 95% confidence interval for the proportion of graduates who found a job within one year of completing their undergraduate degree at this university, and interpret it in the context of the data.(d) What does "95% confidence" mean?(e) Now calculate a 99% confidence interval for the same parameter and interpret it in the context of the data.(f) Compare the widths of the 95% and 99% confidence intervals. Which one is wider? Explain.(Please show work for all problems, thank you)

Answer 1

a) The parameter of interest is p who represent the proportion of graduates from this university who found a job within one year after graduating, and the estimated value is:

`\hat p=(348)/(400)=0.87`

b)
`np=400*0.87=348>10`

`n(1-p)=400(1-0.87)=52>10`

So both conditions are satisifed and we can construct the confidence interval.

c) The 95% confidence interval would be given (0.837;0.903).

We are confident (95%) that that the true proportion of graduates that found jobs is between 0.837 and 0.903

d) On this case that the 95% of the selected random samples will produce a 95% confidence interval that includes the true proportion of interest.

e) The 99% confidence interval would be given (0.827;0.913).

We are confident (99%) that that the true proportion of graduates that found jobs is between 0.827 and 0.913

f) The width for the 95% interval is 0.903-0.837=0.066, and for the 99% interval 0.913-0.827=0.086. And we see that the 99% is wider since we have more confidence that the true parameter of interest would be on the range provided.

Explanation:

Data given and notation

n=400 represent the random sample taken

X=348 represent the number of graduates that found jobs in the sample

`\hat p=(348)/(400)=0.87` estimated proportion of graduates that found jobs

`\alpha` represent the significance level

z would represent the statistic to calculate the confidence interval

p= population proportion of graduates that found jobs

(a) Describe the population parameter of interest. What is the value of the point estimate of this parameter?

The parameter of interest is p who represent the proportion of graduates from this university who found a job within one year after graduating, and the estimated value is:

`\hat p=(348)/(400)=0.87`

(b) Check if the conditions for constructing a confidence interval based on these data are met.

`np=400*0.87=348>10`

`n(1-p)=400(1-0.87)=52>10`

So both conditions are satisifed and we can construct the confidence interval.

(c) Calculate a 95% confidence interval for the proportion of graduates who found a job within one year of completing their undergraduate degree at this university, and interpret it in the context of the data.

The confidence interval would be given by this formula

`\hat p \pm z_(\alpha/2) \sqrt{(\hat p(1-\hat p))/(n)}`

For the 95% confidence interval the value of
`\alpha=1-0.95=0.05` and
`\alpha/2=0.025`, with that value we can find the quantile required for the interval in the normal standard distribution.

`z_(\alpha/2)=1.96`

And replacing into the confidence interval formula we got:

`0.87 - 1.96 \sqrt{(0.87(1-0.87))/(400)}=0.837`

`0.87 + 1.96 \sqrt{(0.87(1-0.87))/(400)}=0.903`

And the 95% confidence interval would be given (0.837;0.903).

We are confident (95%) that that the true proportion of graduates that found jobs is between 0.837 and 0.903

(d) What does "95% confidence" mean?

On this case that the 95% of the selected random samples will produce a 95% confidence interval that includes the true proportion of interest.

(e) Now calculate a 99% confidence interval for the same parameter and interpret it in the context of the data

For the 99% confidence interval the value of
`\alpha=1-0.99=0.01` and
`\alpha/2=0.005`, with that value we can find the quantile required for the interval in the normal standard distribution.

`z_(\alpha/2)=2.58`

And replacing into the confidence interval formula we got:

`0.87 - 2.58 \sqrt{(0.87(1-0.87))/(400)}=0.827`

`0.87 + 2.58 \sqrt{(0.87(1-0.87))/(400)}=0.913`

And the 99% confidence interval would be given (0.827;0.913).

We are confident (99%) that that the true proportion of graduates that found jobs is between 0.827 and 0.913

(f) Compare the widths of the 95% and 99% confidence intervals. Which one is wider?

The width for the 95% interval is 0.903-0.837=0.066, and for the 99% interval 0.913-0.827=0.086. And we see that the 99% is wider since we have more confidence that the true parameter of interest would be on the range provided.