Question

The caloric consumption of

43

adults was measured and found to average

2 ,155

Assume the population standard deviation is

263

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.	93 %
b.	96 %
c.	97%

a. The

93 %

confidence interval has a lower limit of

nothing

and an upper limit of

nothing.

​(Round to one decimal place as​ needed.)

b. The

96 %

confidence interval has a lower limit of

nothing

and an upper limit of

nothing.

​(Round to one decimal place as​ needed.)

c. The

97 %

confidence interval has a lower limit of

nothing

and an upper limit of

nothing.

​(Round to one decimal place as​ needed.)

Answer 1

Solution :

Given that,

= 2155

= 263

n = 43

a ) At 93% confidence level the z is ,

  = 1 - 93% = 1 - 0.93 = 0.07

/ 2 = 0.07 / 2 = 0.035

Z_/2 = Z_0.035 = 1.812

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

= 1.812 * ( 263 / 43 )

= 72.7

At 93% confidence interval estimate of the population mean is,

- E < < + E

2155 - 72.7< < 2155 + 72.7

2082.3 < < 2227.2

lower limit = 2082.3

Upper limit = 2227.2

b ) At 96% confidence level the z is ,

  = 1 - 96% = 1 - 0.96 = 0.04

/ 2 = 0.04 / 2 = 0.02

Z_/2 = Z_0.02 = 2.054

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

= 2.054 * ( 263 / 43 )

= 83.4

At 96% confidence interval estimate of the population mean is,

- E < < + E

2155 - 83.4< < 2155 + 83.4

2072.6 < < 2238.4

lower limit = 2072.6

Upper limit = 2238.4

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 * ( 263 / 43 )

= 83.0

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

- E < < + E

2155 - 83.0< < 2155 + 83.0

2072.0 < < 2227.2

lower limit = 2072.0

Upper limit = 2238.0