Mathematics: Confidence Intervals + Mean exam question.?

Question

I'd studying this squestion at the moment for my exam next week. I'd really like an example to work form.

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A third level college employs several lecturing staff on a part-time basis. A random sample of 40 part-time staff showed a mean weekly earning of €375 per person and a Standard Deviation of €83.

i) Calculate a 95% confidence interval for the mean weekly earning of all part-time staff.

ii) A survey of 180 staff members showed that 52 of them were female. Use this information to estimate, with 99 percent confidence, the proportion of all staff that is female. Interpret your result.

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That's the question. If it's OK, I'd also like an explanation of what "confidence intervals" are.. Or maybe some direction on where to find out. Also, interpreting my result? What does that entail?

Answer 1

i)

Since the population standard deviation is unknown, we use the t-distribution with 40-1 = 39 degrees of freedom. The critical value that corresponds to 95 % confidence level (close to 39) is 2.021.

Sample mean 375

Standard deviation = 63

Standard error of mean = sd / sqrt(n)

SE = 63/6.3246

Standard error of mean 9.9612

Confidence limits 375-(9.9612)(2.021)

and 375+(9.9612)(2.021)

(354.8685, 395.1315) is the confidence interval.

This means there is a 95 % probability that the true mean of all of the part-time staff (which we don't know) lies anywhere between 354.87 and 395.13

ii)We assume Normal distribution in this case and the critical value that corresponds to 99 % is 2.57.

Varaince of proportion = p*(1-p)/n

52/180=0.288889

= 0.288889(0.711111)/180 =0.0011413

S.D. of p is 0.0338

Confidence limits:

phat-zval*sd = 0.2889 - (2.57)(0.033783)

phat-zval*sd = 0.2889 + (2.57)(0.033783)

99 % Confidence limits are ( 0.2021 , 0.3757 )

This means there is a 99 % probability that the true proportion of female staff  (which we don't know) lies anywhere between 0.2021 and 0.3757.