Question

Suppose that the lengths, in inches, of adult corn snakes are normally distributed with an unknown mean and standard deviation. A random sample of 38 snakes is taken and gives a sample mean of 53 inches and a sample standard deviation of 6 inches. Find a 95% confidence interval estimate for the population mean using the Student's t-distribution.

df...3435363738t0.10…1.3071.3061.3061.3051.304t0.05…1.6911.6901.6881.6871.686t0.025…2.0322.0302.0282.0262.024t0.01…2.4412.4382.4342.4312.429t0.005…2.7282.7242.7192.7152.712

Use the portion of the table above or a calculator.

Answer 1

solution

Given that,

= 53

s =6

n = 38

Degrees of freedom = df = n - 1 = 38- 1 = 37

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t /2,df = t0.025,37 =2.026 ( using student t table)

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

= 2.026 * (6 / 38) = 1.9720

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

- E < < + E

53 - 1.9720< < 53+ 1.9720

51.0280< < 54.9720

( 51.0280, 54.9720 )