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MakcuM [25]
3 years ago
11

A sample of 900 college freshmen were randomly selected for a national survey. Among the survey participants, 372 students were

pursuing liberal arts degrees. The sample proportion is 0.413.What is the margin of error for a 99% confidence interval for this sample? What is the lower endpoint for the 99% confidence interval?
Mathematics
1 answer:
Rufina [12.5K]3 years ago
7 0

Answer:

a) ME=2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.0423

b) 0.413 - 2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.371

Step-by-step explanation:

1) Data given and notation  

n=900 represent the random sample taken    

X=372 represent the students were pursuing liberal arts degrees

\hat p=\frac{372}{900}=0.413 estimated proportion of students were pursuing liberal arts degrees

\alpha=0.01 represent the significance level

z would represent the statistic (variable of interest)    

p_v represent the p value (variable of interest)    

p= population proportion of students were pursuing liberal arts degrees

2) Solution to the problem

The confidence interval would be given by this formula

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

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

The margin of error is given by:

ME=z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

If we replace we have:

ME=2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.0423

And replacing into the confidence interval formula we got:

0.413 - 2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.371

0.413 + 2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.455

And the 99% confidence interval would be given (0.371;0.455).

We are confident (99%) that about 37.1% to 45.5% of students were pursuing liberal arts degrees.

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