Question

The weight of gravel, in kilograms, collected by a power shovel may be modeled by a normal distribution with unknown mean, µ, and standard deviation 50 kg. The weights of a random sample of 20 collections have a mean of 1030 kg. Answer the following questions round all FINAL answers to two decimal places.

(a) What is the point estimate of the mean weight of gravel that the shovel can collect?

(b) Calculate the margin of error for a 95% confidence interval estimate of µ

(c) State the 95% confidence interval for the mean, µ ( , )

(d) What sample size is needed so that the error is no more than 10 kg? n= power shovels

Answer 1

Solution :

Given that,

a) Point estimate = sample mean = = 1030 kg

sample standard deviation = s = 50 kg

sample size = n = 20

Degrees of freedom = df = n - 1 = 20 - 1 = 19

b) At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

t/2,df = t0.025,19 = 2.093

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

= 2.093 * (50/ 20)

Margin of error = E = 23.40

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

  ± E  

= 1030 ± 23.40

= ( 1006.60, 1053.40 )

d) margin of error = E = 10

sample size = n = [t/2,df* s / E]2

n = [2.093 * 50 / 10 ]2

n = 27.37

Sample size = n = 28