Question

(5) In a random sample of 21 ​people, the mean commute time to work was 31.5 minutes and the standard deviation was 7.3 minutes. Assume the population is normally distributed and use a​ t-distribution to construct a 80​% confidence interval for the population mean μ. What is the margin of error of μ​? Interpret the results.

(8) Use the standard normal distribution or the​ t-distribution to construct a 99​% confidence interval for the population mean. Justify your decision. If neither distribution can be​ used, explain why. Interpret the results. In a random sample of 17 mortgage​ institutions, the mean interest rate was 3.69​% and the standard deviation was 0.38​%. Assume the interest rates are normally distributed.

(3)

Construct the indicated confidence interval for the population mean μ using the​ t-distribution. Assume the population is normally distributed. c= 0.99​, overbar x=12.9​, s= 4.0​,n=5 round to one decimal place

Answer 1

Solution :

5 ) Given that,

= 31.5

s = 7.3

n = 21

Degrees of freedom = df = n - 1 = 21 - 1 = 20

At 80%confidence level the t is ,

= 1 - 80% = 1 - 0.80 = 0.20

/ 2 = 0.20 / 2 = 0.10

t _/2,df = t_0.10,20 =1.325

Margin of error = E = t_/2,df * (s /n)

= 1.325 * (7.3 / 21)

= 2.1

Margin of error = 2.1

The 80% confidence interval estimate of the population mean is,

- E < < + E

31.5 - 2.1 < < 31.5 + 2.1

29.4 < < 32.6

(29.4, 32.6 )

8 ) Given that,

= 0.0369

s = 0.0038

n = 17

Degrees of freedom = df = n - 1 = 17 - 1 = 16

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t _/2,df = t_0.005,16 =2.921

Margin of error = E = t_/2,df * (s /n)

= 2.921 * (0.0038 / 17

= 0.0027

Margin of error = 0.0027

The 99% confidence interval estimate of the population mean is,

- E < < + E

0.0369 - 0.0027 < < 0.0369 + 0.0027

0.0342 < < 0.0396

(0.0342, 0.0396 )

3 ) Given that,

= 12.9

s = 4.0

n = 5

Degrees of freedom = df = n - 1 = 5 - 1 = 4

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t _/2,df = t_0.005,4 =4.604

Margin of error = E = t_/2,df * (s /n)

= 4.604 * (4.0 / 25)

= 8.2

Margin of error = 8.2

The 99% confidence interval estimate of the population mean is,

- E < < + E

12.9 - 8.2 < < 12.9 + 8.2

4.7< < 21.1

(4.7, 21.1 )