Question

The delivery times for a distributing site are normally distributed with an unknown population mean and standard deviation. If a random sample of 25 deliveries is taken to estimate the mean delivery time, what t -score should be used to find a 98% confidence interval estimate for the population mean?

df t 0.10 t 0.05 t 0.025 t 0.01 t 0.005 ... … … … … … 21 1.323 1.721 2.080 2.518 2.831 22 1.321 1.717 2.074 2.508 2.819 23 1.319 1.714 2.069 2.500 2.807 24 1.318 1.711 2.064 2.492 2.797 25 1.316 1.708 2.060 2.485 2.787

Use the portion of the table above or a calculator. If you use a calculator, round your answer to three decimal places.

Answer 1

Given ,

Sample size = n = 25

Confidence level = 98% = 0.98

Significance level = =1 - 0.98 = 0.02

Degrees of freedom = df = n - 1 = 25 - 1 = 24

We have to find two tailed t-score for given confidence level.

Using t-score table,

t-score = 2.492

OR

Using Excel function ,

=TINV( , df ) ---> this function returns two tailed inverse of t distribution.

Here ,   = 0.02 , df = n - 1 = 25- 1 = 24

=TINV( 0.02 , 24 ) = 2.492