Question

Please answer all parts

1. Confidence Interval Given. Assume I created a 95% confidence interval for the mean hours studied for a test based on a random sample of 64 students. The lower bound of this interval was 3.1416 and the upper bound was 18.6282. Assume that when I created this interval I knew the population standard deviation. Keep all decimals in your calculations.

a) Calculate the width of the interval.

(b) Calculate the margin of error for the interval.

(c) Calculate the center of the interval.

(d) What is the sample mean?

(e) What is the z ∗ (or zα/2) used? (

f) Calculate the population standard deviation. [Do not use the Empirical Rule.]

Answer 1

n = number of student = 64

lower bound = 3.1416

Upper bound = 18.6282

a) Width of confidence interval = Upper bound - lower bound = 18.6282 - 3.1416 = 15.4866

Answer :

Width of confidence interval   = 15.4866

b)

Margin of error for the interval : M.E

Answer :

Margin of error = 7.7433

c) Center of the confidence interval

Center of the confidence interval is nothing but the sample mean

Answer :-

Center of the confidence interval = 10.8849

d)

We use formula for sample mean

Answer :-

Sample mean = 10.8849

e )

Now we find the critical value

c = level of confidence = 0.95

Now we use Z table to find the z critical value

We search for the area 0.9750 inside the body of the table

Z score = Z critical value = row headed number for 0.9750 + column headed number for 0.9750

Z score = 1.9 + 0.06 = 1.96

Z = Z_{(α /2)} = 1.96

Answer:-

Z_{(α /2)} = 1.96

f )

We use formula for margin of error

We have calculated above

M.E = 7.7433

Z_{(critical value )} = 1.96

n = sample size = 64

Answer:-

population standard deviation = 31.60530612

I hope this will help you :)