A random sample of 78 students was interviewed, and 59 students said that they w

Question

A random sample of 78 students was interviewed, and 59 students said that they would vote for Mandy Milcovich as student body who will vote for Mandy. Find a (a) Let p represent the proportion of all students at this college point estimate for p and for q (b) Find 90% confidence interval for p and g. Use 3 decimal places for each and the E term. (e) What assumptions are required for the calculation of part (b? Are these conditions satisfied? (d) How many more students should be included in the sample to be 90% sure that a point estimate will be within a distance of 0.05 from p? 4. A random sample of 53 students was asked for the number of semester hours they are taking this semester. The sample standard deviation was found to be s 4.7 semester hours. How many more students should be included in the sample to be 99% sure that the sample mean is within semester hour ofthe population mean H for all students at this college?

Explanation / Answer

59 out of 78 vote for Mandy

a) p = 59/78

= 0.756

Point estimate of p = mean = np

Mean = 78*0.756 = 59

b) at 90% interval Z = 1.645

= p +/- Z * sqrt (p(1-p)/n)

= 0.756 +/- 1.645*sqrt (0.756*0.244/78)

= 0.456 +/- 0.0799

= 0.6761 to 0.8359

c) Z test assumes value of sigma is unknown

Also Z test assumes that sample is large enough to be a normally distributed

Here based on the assumption of normality we does CI.

d)

New P = 0.05 + p

= 0.806

= 62.8 = 63

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