Question

Let the following sample of 8 observations be drawn from a normal population with unknown mean and standard deviation: 24, 23, 13, 22, 14, 17, 29, 19.

a. Calculate the sample mean and the sample standard deviation. (Round intermediate calculations to at least 4 decimal places. Round "Sample mean" to 3 decimal places and "Sample standard deviation" to 2 decimal places.)

b. Construct the 95% confidence interval for the population mean. (Round "t" value to 3 decimal places and final answers to 2 decimal places.)

c. Construct the 99% confidence interval for the population mean. (Round "t" value to 3 decimal places and final answers to 2 decimal places.)

Answer 1

Solution :

Given that,

a)

Point estimate = sample mean = = Xi / n

= ( 24+23+13+22+14+17+29+19 )/8

= 161 / 8

= 20.125

sample standard deviation = s = ( X - ) / n-1

= (24-20.125)+(23-20.125)+(13-20.125)+(22-20.125)+(14-20.125)+(17-20.125)+(29-20.125)+(19-20.125) / 8 - 1

= 37.87 / 7

= 5.41

sample size = n = 8

Degrees of freedom = df = n - 1 = 7

b)

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,7 = 2.365

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

= 2.365 * ( 5.41/ 8)

= 4.52

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

- E < < + E

20.125 - 4.52 < < 20.125 + 4.52

15.61 < < 24.66

(15.61 , 24.66)

c)

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t _/2,df = t_0.005,7 = 3.499

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

= 3.499 * ( 5.41/ 8)

= 6.69

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

- E < < + E

20.125 - 6.69 < < 20.125 +6.69

15.61 < < 26.82

(15.61 , 26.82)