Question

The daily intakes of milk (in ounces) for ten randomly selected people were:

26.0    13.0    13.5    14.8    19.1 13.6    21.6    30.8    15.8    14.1

Find a 99% confidence interval for the population variance. Assume that the distribution of ounces of milk intake daily is normally distributed.

What is the confidence interval?

lower limit =Incorrect answer:

and the upper limit =Incorrect answer:

Answer 1

ANSWER:

Given that,

The daily intakes of milk (in ounces) for ten randomly selected people were:

26.0

13.0

13.5

14.8

19.1

13.6

21.6

30.8

15.8

14.1

= x / n = (26.0+13.0+13.5+14.8+19.1+13.6+21.6+30.8+15.8+14.1) / 10 = 182.3/10 = 18.23

s^2 = (x-)^2 / (n-1)

= ((26.0-18.23)^2+(13.0-18.23)^2+(13.5-18.23)^2+(14.8-18.23)^2+(19.1-18.23)^2+(13.6-18.23)^2+(21.6-18.23)^2+(30.8-18.23)^2+(15.8-18.23)^2+(14.1-18.23)^2) / (10-1)

= (60.3729+27.3529+22.3729+11.7649+0.7569+21.4369+11.3569+158.0049+5.9049+17.0569) / (9)

= 336.381 /9

= 37.3757

Sample size = n = 10

Degree of freedom = df = n-1 = 10-1 = 9

c = 99% = 99/100 = 0.99

= 1-c = 1-0.99 = 0.01

/2 = 0.01/2 = 0.005

1-(/2 ) = 1-(0.01/2) = 0.995

Critical values,

= = 1.7349 and

= = 23.5893

Confidence interval

(n-1) / < < (n-1) /

(10-1)37.3757 / 23.5893< < (10-1)37.3757 / 1.7349

14.2599 < < 193.8909

The confidence interval is 14.2599 ounces < < 193.8909 ounces

lower limit =14.2599 ounces

upper limit =193.8909 ounces

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