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Consider college officials in admissions registration, counseling, financial aid campus ministry, food services, and so on. How much money do these people make each year? Suppose you read in your local newspaper that 45 officials in student services earn an average of $50,340 each year. Assume that the standard deviation is $10,780 for salaries of college officials and student services. Find a 90% confidence interval for the population mean salaries of such personnel. Round your answer to the nearest dollar and don;t forget to use the $ sign.

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Since sample size is > 40, we use the z-score
in calculating for the confidence interval.

The formula is given as:

Confidence Interval = X ± z * σ / sqrt (n)

Where,

X = mean = $50,340

z = z-score which is taken from standard distribution
tables at 90% confidence interval = 1.645

σ
= standard deviation = $10,780

n = sample size = 45

Substituting to the equation:

Confidence Interval = 50,340 ± 1.645 * 10,780 / sqrt (45)

Confidence Interval = 50,340 ± 1,607

Confidence Interval = $48,733 to $51,947

Therefore the salary range of the personnel is $48,733 to $51,947.

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There are few industries that have such a wide range of salaries as the entertainment industry. salaries spread from those for the "dollar a holler" commercial voiceovers to the multi-million dollar movie stars. obviously there is little ___________ in hollywood.

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There are few industries that have such a wide range of salaries as the entertainment industry. salaries spread from those for the “dollar a holler” commercial voiceovers to the multi-million dollar movie stars. obviously there is little ___________ in hollywood.

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