In a simple random sample of 380 young people, 85% had earned a high school diploma. Complete the following parts (Round all answers to 2 decimal places.)

a) What is the standard error for this estimate of the percentage of all young people who earned a high school diploma? (as a percentage)
Answer: __________

b) Find the margin of error, using a 95% confidence level, for estimating the percentage of all young people who earned a high school diploma. (as a percentage)
Answer: __________

c) Find the 95% confidence interval for the percentage of all young people who earned a high school diploma.
Answer: __________ % to __________ %



Answer :

Let's solve this problem step by step:

### Part a) Calculate the Standard Error for the Estimate

To find the standard error for the proportion, we use the formula for the standard error of a proportion:
\
[tex]\[ \text{Standard Error (SE)} = \sqrt{\frac{p(1 - p)}{n}} \][/tex]

where:
- [tex]\( p \)[/tex] is the sample proportion (85% or 0.85),
- [tex]\( n \)[/tex] is the sample size (380).

Putting the values into the formula:
\
[tex]\[ \text{SE} = \sqrt{\frac{0.85 \times (1 - 0.85)}{380}} \][/tex]

Converting the standard error to a percentage:
\
[tex]\[ \text{SE as percentage} = \text{SE} \times 100 \][/tex]

Therefore, the standard error as a percentage is:
\
[tex]\[ \text{Standard Error} \approx 1.83\% \][/tex]

### Part b) Calculate the Margin of Error Using a 95% Confidence Level

The margin of error (ME) can be found using the z-score for a 95% confidence level, which is approximately 1.96. The formula is:
\
[tex]\[ \text{Margin of Error (ME)} = z \times SE \][/tex]

where [tex]\( z \)[/tex] is the z-score and [tex]\( SE \)[/tex] is the standard error.

In percentage terms:
\
[tex]\[ \text{ME as percentage} = z \times (\text{SE as percentage}) \][/tex]

Therefore, the margin of error is:
\
[tex]\[ \text{Margin of Error} \approx 3.59\% \][/tex]

### Part c) Calculate the 95% Confidence Interval

The confidence interval is calculated by taking the sample proportion and adding and subtracting the margin of error from it.

The lower limit of the confidence interval (CI) is:
\
[tex]\[ \text{Lower limit} = (p - \text{ME}) \times 100 \][/tex]

The upper limit of the confidence interval (CI) is:
\
[tex]\[ \text{Upper limit} = (p + \text{ME}) \times 100 \][/tex]

Using the margin of error from Part b:
\
[tex]\[ \text{Lower limit} \approx 85\% - 3.59\% = 81.41\% \][/tex]
\
[tex]\[ \text{Upper limit} \approx 85\% + 3.59\% = 88.59\% \][/tex]

Thus, the 95% confidence interval for the percentage of all young people who earned a high school diploma is:
\
[tex]\[ 81.41\% \text{ to } 88.59\% \][/tex]

### Summary of Results
a) Standard Error: [tex]\( 1.83\% \)[/tex]

b) Margin of Error: [tex]\( 3.59\% \)[/tex]

c) 95% Confidence Interval: [tex]\( 81.41\% \text{ to } 88.59\% \)[/tex]