Question

A news report states that the 99% confidence interval for the mean number of daily calories consumed by participants in a medical study is (2070, 2290). Assume the population distribution for daily calories consumed is normally distributed and that the confidence interval was based on a simple random sample of 22 observations. Calculate the sample mean, the margin of error, and the sample standard deviation based on the stated confidence interval and the given sample size. Use the t distribution in any calculations and round non-integer results to 4 decimal places.

Answer 1

Solution :

Given a 99% confidence interval for mean is (2070, 2290)

n = 22

Upper limit = 2290

Lower limit = 2070

We know that , the confdience interval for mean is nothing but

sample mean margin of error

i.e.       /2,d.f. * ( / n )

Upper limit = sample mean +  margin of error

Lower limit = sample mean -  margin of error

1) Sample mean = (Upper limit + Lower limit)/2

= ( 2290 + 2070)/2

= 2180

Sample mean = 2180

2)

Margin of error = (Upper limit - Lower limit)/2

= ( 2290 - 2070)/2

= 110

Margin of error = 110

3)Here , n =22 given

d.f. = n -1 = 21

c = 99% = 0.99 confidence level given

= 1 - c = 1 - 0.99 = 0.01

  /2 = 0.01 2 = 0.005

    =    =  _0.00521 = 2.831 ...use t table

The formula for margin of error is

Margin of error =  /2,d.f. * ( / n )

110 =  2.831 * (s/22)

s = (110 *  22 ) /2.831

= 182.2486

sample standard deviation = 182.2486