Question

Use the standard normal distribution or the​ t-distribution to construct a 99​% 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 26.1 and the standard deviation was 6.06.

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

Answer 1

solution:

conditons : when we use z distribution 1.sample size greater than 30

2.population standard deviation given

condition:when we use t distribution 1.sample size less than 30

2.sample standard deviation given

Solution :

Given that,

Point estimate = sample mean = $\bar x$ = 26.1

Population standard deviation =   $\sigma$ = 6.06

Sample size = n =48

At 99% confidence level the z is ,

$\alpha$  = 1 - 99% = 1 - 0.99 = 0.01

$\alpha$ / 2 = 0.01 / 2 = 0.005

Z$\alpha$/2 = Z0.005 = 2.576   ( Using z table )

Margin of error = E = Z$\alpha$/2* ($\sigma$ /$\sqrt$n)

= 2.576 * ( 6.06/ $\sqrt$ 48)

= 2.25

At 99% confidence interval is,

$\bar x$ - E < $\mu$ < $\bar x$ + E

26.1-2.25< $\mu$ < 26.1+2.25

23.85< $\mu$ < 28.35

(23.85 , 28.35)

here we use z distribution (standard normal )