Question

The caloric consumption of 44 adults was measured and found to average 2133. Assume the population standard deviation is 270 calories per day. Construct confidence intervals to estimate the mean number of calories consumed per day for the population with the confidence levels shown below.

a. 91%

b. 95%

c. 97%

Answer 1

Solution :

Given that,

= 2133

= 270

n = 44

a ) At 91% confidence level the z is ,

  = 1 - 91% = 1 - 0.91 = 0.09

/ 2 = 0.09 / 2 = 0.045

Z_/2 = Z_0.045 = 1.695

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

= 1.695 * ( 270 / 44)

= 69

At 91% confidence interval estimate of the population mean is,

- E < < + E

2133 - 69 < < 2133 + 69

2064 < < 2202

( 2064 , 2202 )

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 = 1.960

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

= 1.960 * ( 270 / 44)

= 80

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

- E < < + E

2133 - 80 < < 2133 + 80

2053 < < 2213

( 2053 , 2213 )

c ) At 97% confidence level the z is ,

  = 1 - 97% = 1 - 0.97 = 0.03

/ 2 = 0.03 / 2 = 0.015

Z_/2 = Z_0.015 = 2.170

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

= 2.170 * ( 270 / 44)

= 88

At 97% confidence interval estimate of the population mean is,

- E < < + E

2133 - 88 < < 2133 + 88

2045 < < 2221

( 2045 , 2221 )