Question

The table below contains data on sugar (gms/serving) in samples of cereal geared toward children.

10	14	12	9	13	13	13
11	12	15	9	10	11	3
6	12	15	12	12

Assuming the population is known to be normally distributed, find the 95% confidence interval.  Give answers to 2 decimal places.

Using your answer to part a, give the result in terms of mean ± margin of error.  Give answer to 2 decimal places.

Answer 1

From the data given, we first calculate the mean and the standard deviation.

n = 19

Mean = (x)/n

= 212/19

= 11.158

Sample standard deviation = [(1/n-1)(x - mean)2]

= (156.526 / 18)

= 2.948

Therefore, = 11.158 and s = 2.948

We use the t distribution here.

degrees of freedom = n -1 = 18

For df 18 and 95% confidence level, t = 2.1

The confidence interval is = t * s/n

= 11.158 1.42

= ( 9.74, 12.58 )

In terms of mean ± margin of error , the answer is 11.158 1.42.