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.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

Answer with Step-by-step explanation:

Since we have given

n = 16

a) Describe the sampling distribution for the sample mean.

The sampling distribution of the mean is the population's mean from where the items are sampled.

It would be $\mu$

b. What is the standard error?

$s=\dfrac{\sigma}{\sqrt{n}}\\\\s=\dfrac{14}{\sqrt{16}}\\\\s=\dfrac{14}{4}\\\\s=3.5$

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

Degrees of freedom = v - 1= 16-1 = 15

Using t-table, 1.753 cuts 5% level of significance in each tail.

so, Margin of error would be

$1.753\times 3.5\\\\=6.1355$

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

So, 90% confidence interval would be

$110+6.136\\\\=116.136\\\\and\\\\110-6.136\\\\=103.86$

(103.86,116.136)