Question

1. For a sample, the mean is 34.2, standard deviation is 5.3, and the sample size is 35.

a.) what is the point estimate for the population mean?

b.) compute a 95% confidence interval about the mean for the data (use t table)

c.) compute a 99% confidence interval about the mean (use t table)

2. a poll asked 25 americans "during the past year how many books did you read?" the mean # was 18.8 books, and stand deviation was 19.8 books.

a. what is the point estimate for the number of books americans read? (use t table). also construst a 99% confidence interval for the mean # of books.

Answer 1

Solution(1)
Given in the question
No. of sample = 35
Sample mean = 34.2
Sample standard deviation = 5.3
Solution(1a)
Point estimate for the population mean = Sample mean = 34.2
Solution(1b)
95% confidence interval can be calculated as
Mean +/- talpha/2*S/sqrt(n)
here alpha = 0.05, alpha/2 = 0.05/2 = 0.025 and df = 35-1 =34, so talpha/2 from t table is 2.03
So 95% confidence interval is
34.2 +/- 2.03*5.3/sqrt(35)
34.2 +/- 1.82
So 95% confidence interval is 32.38 to 36.02
Solution(1b)
99% confidence interval can be calculated as
Mean +/- talpha/2*S/sqrt(n)
here alpha = 0.01, alpha/2 = 0.01/2 = 0.005 and df = 35-1 =34, so talpha/2 from t table is 2.73
So 95% confidence interval is
34.2 +/- 2.73*5.3/sqrt(35)
34.2 +/- 2.45
So 99% confidence interval is 31.75 to 36.65
Solution(2)
Given in the question
No. of sample = 25
Sample mean = 18.8
Sample standard deviation = 19.8
Solution(a)
Point estimate for the number of books americans read = 18.8
99% confidence interval can be calculated as
Mean +/- talpha/2*S/sqrt(n)
here alpha = 0.01, alpha/2 = 0.01/2 = 0.005 and df = 25-1 =24, so talpha/2 from t table is 2.8
So 95% confidence interval is
18.8 +/- 2.8*19.8/sqrt(25)
18.8 +/- 11.09
So 99% confidence interval is 7.71 to 29.89