Use the standard normal distribution or the t-distribution to construct a 95% co

Question

Use the standard normal distribution or the t-distribution to construct a 95% confidence interval for the population mean. Justify your decision. If neither distribution can be used, explain why. Interpret the results.

In a random sample of 48 people, the mean body mass index (BMI) was 26.4 and the standard deviation was 6.01.

Which distribution should be used to construct the confidence interval? Choose the correct answer below.

A. Use a t-distribution because the sample is random, the population is normal, and is ?

unknown.

B. Use a normal distribution because the sample is random, n ? 30, and ?

is known.

C. Use a normal distribution because the sample is random, the population is normal, and ?

is known.

D. Use a t-distribution because the sample is random, n ? 30, and ?

is unknown.

E. Neither a normal distribution nor a t-distribution can be used because either the sample is not

random, or n < 30, and the population is not known to be normal.

Select the correct choice below and, if necessary, fill in any answer boxes to complete your choice.

A. The 95 % confidence interval is

( _____,_______ ).

(Round to two decimal places as needed.)

B. Neither distribution can be used to construct the confidence interval.

Interpret the results. Choose the correct answer below.

A. With 95% confidence, it can be said that the population mean BMI is between the bounds of

the confidence interval.

B. If a large sample of people are taken approximately 95% of them will have a BMI between

the bounds of the confidence interval.

C. It can be said that 95 % of people have a BMI between the bounds of the confidence interval.

D. Neither distribution can be used to construct the confidence interval.

Explanation / Answer

We use t distribution because the sample is random the population is normal and population SD is unknown.

A)

Give n = 48

Df = n-1 = 47

Sample mean = xbar = 26.4

Sample SD = s = 6.01

Los be Alfa = 0.05 then tcri = 2.012

95% confidence interval is

Xbar +/- ME

= 26.4 +/- (2.012*6.01/?48)

= 26.4 +/- 1.745

= (24.655 , 28.145)

B) we are 95% confident that the population mean BMI is between the bounds of 24.66 and 28.15

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