Question

A sample of 1600 observations from a normal distribution has sample mean 50 and sample standard deviation 10.

a. What is the point estimate for the population mean of X?

b. Write a 95% confidence interval for the population mean of X (Use the t table to obtain the critical value and round to two decimal places). ( , )

c. Write a 99% confidence interval for the population mean of X (Use the t table to obtain the critical value and round to two decimal places). ( ,   )

Answer 1

Solution :

Given that,

= 50

s =10

n =1600

Degrees of freedom = df = n - 1 = 1600- 1 =1599

a ) At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

  / 2= 0.05 / 2 = 0.025

t /2,df = t0.025,1599 = 1.96 ( using student t table)

Margin of error = E = t/2,df * (s /n)

=1.96 * (10 / 1600)

= 0.4900

The 95% confidence interval estimate of the population mean is,

- E < < + E

50 - 0.4900 < <50 + 0.4900

49.5100 < < 50.4900

( 49.5100, 50.4900)

(B)

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t /2  df = t0.005, 1599= 2.58 ( using student t table)

Margin of error = E = t/2,df * (s /n)

=2.58 * (10 / 1600) = 0.6450

The 99% confidence interval estimate of the population mean is,

- E < < + E

50 - 0.6450 < < 50+ 0.6450

49.3550 < < 50.6450

( 49.3550 ,50.6450)