Question

Use the sample information x⎯⎯x = 36, σ = 7, n = 16 to calculate the following confidence intervals for μ assuming the sample is from a normal population.

(a) 90 percent confidence. (Round your answers to 4 decimal places.)

The 90% confidence interval is from  to

(b) 95 percent confidence. (Round your answers to 4 decimal places.)

The 95% confidence interval is from  to

(c) 99 percent confidence. (Round your answers to 4 decimal places.)

The 99% confidence interval is from  to

(d) Describe how the intervals change as you increase the confidence level.

	The interval gets narrower as the confidence level increases.
	The interval gets wider as the confidence level decreases.
	The interval gets wider as the confidence level increases.
	The interval stays the same as the confidence level increases.

Answer 1

Solution :

Given that,

= 36

= 7

n = 16

(a)

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

Margin of error = E = Z_/2* ( /n)

= 1.645 * ( 7 / 16)

= 2.8786

At 90% confidence interval estimate of the population mean is,

- E < < + E

36 - 2.8786 < < 36 + 2.8786

33.1214 < < 38.8786

The 90% confidence interval is from  to : 33.1214 to 38.8786

(b)

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Margin of error = E = Z_/2* ( /n)

= 1.96 * ( 7 / 16)

= 3.43

At 95% confidence interval estimate of the population mean is,

- E < < + E

36 - 3.43 < < 36 + 3.43

32.57 < < 39.43

The 95% confidence interval is from  to : 32.57 to 39.43

(c)

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

Margin of error = E = Z_/2* ( /n)

= 2.576 * ( 7 / 16)

= 4.508

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

- E < < + E

36 - 4.508 < < 36 + 4.508

31.492 < < 40.508

The 99% confidence interval is from  to : 31.492 to 40.508

d)

The interval gets wider as the confidence level increases.