Question

A small business ships specialty homemade candies to anywhere in the world. Past records indicate that the weight of orders is normally distributed. Suppose a random sample of 16 orders is selected and each is weighed. The sample mean was found to be 110 grams with a standard deviation of 14 grams. Show your work

A. Describe the sampling distribution for the sample mean.

B. What is the standard error?

C.. For 90% confidence, what is the margin of error?

D. Based on the sample results, create the 90% confidence interval and interpret.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 110

sample standard deviation = s = 14

sample size = n = 16

Degrees of freedom = df = n - 1 = 16 - 1 = 15

A) The sampling distribution for the sample mean is normal.

B) SE = s / n = 14 / 16 = 3.5

C) At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

t/2,df = t0.05,15 = 1.753

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

= 1.753 * ( 14/ 16)

Margin of error = E = 6.14

D) The 90% confidence interval estimate of the population mean is,

  ± E  

= 110  ± 6.14

= ( 103.86, 116.14 )

We are 90% confident that the true mean of Past records indicate that the weight of orders is between 103.86 and 116.14.