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2 years ago

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A university surveyed recent graduates of the English department for their starting salaries. Four hundred graduates returned th

e survey. The average salary was $25,000. The population standard deviation was $2,500. What is the 95% confidence interval for the mean salary of all graduates from the English department

1 answer:

NeX [460]2 years ago

3 0

Answer:

Step-by-step explanation:

We want to determine a 95% confidence interval for the mean salary of all graduates from the English department.

Number of sample, n = 400

Mean, u = $25,000

Standard deviation, s = $2,500

For a confidence level of 95%, the corresponding z value is 1.96. This is determined from the normal distribution table.

We will apply the formula

Confidence interval

= mean ± z × standard deviation/√n

It becomes

25000 ± 1.96 × 2500/√400

= 25000 ± 1.96 × 125

= 25000 ± 245

The lower end of the confidence interval is 25000 - 245 =24755

The upper end of the confidence interval is 25000 + 245 = 25245

Therefore, with 95% confidence interval, the mean salary of all graduates from the English department is between $24755 and $25245

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The third step would be calculate the standard deviation for the differences, and we got:

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Answer:

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