Question

The Crossett Trucking Company claims that the mean mass of its delivery trucks when they are fully loaded is 2550 kg and the standard deviation is 63 kg. Assume that the population follows the normal distribution. Forty trucks are randomly selected and their masses measured. Within what limits will 98% of the sample means occur? (Round the final answers to the nearest whole number.)

Sample means             to

Answer 1

Solution :

Given that,

= 2550

s = 63

n = 40

Degrees of freedom = df = n - 1 = 40 - 1 = 39

At 98% confidence level the t is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

t /2,df = t0.01,39 =2.426

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

= 2.426 * (63 / 40)

= 24

Margin of error = 24

The 98% confidence interval estimate of the population mean is,

- E < < + E

2550 - 24 < < 2550 +24

2526 < < 2574

(2526, 2574 )