7 A clinic offers a weightloss program A review of its recor

7. A clinic offers a weight-loss program. A review of its records found the following amounts of weight loss in pounds for a random sample of 16 of its clients at the conclusion of a 4-month program:

18 25 16 11 15 20 16 19 28 25 26 31 42 38 35 19

a. Calculate the mean and variance of the sample.

b. Build a 99% confidence interval for the mean weight loss of the people going through this program.

c. If the total number of people who had registered for this weight-loss program had been 100, would your conclusions in (b) above be different? Calculate and explain.

Solution

a)
Mean(x)=24
Standard deviation( sd )=8.914
Variance = S^2 = 79.459
b)
Confidence Interval
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval

Sample Size(n)=16
Confidence Interval = [ 24 ± t a/2 ( 8.914/ Sqrt ( 16) ) ]
= [ 24 - 2.947 * (2.229) , 24 + 2.947 * (2.229) ]
= [ 17.433,30.567 ]
c)
When Sample Size(n)=100
Confidence Interval = [ 24 ± t a/2 ( 8.914/ Sqrt ( 100) ) ]
= [ 24 - 2.626 * (0.891) , 24 + 2.626 * (0.891) ]
= [ 21.659,26.341 ]

It is diffrent, as the reason confidence interval will be vary for every unique value of size (n) .