Question

A meat-processing company in Alberta produces and markets a package of eight small sausage sandwiches. The product is nationally distributed, and the company is interested in knowing the average retail price charged across the country. A random sample of 25 retailers was selected giving a sample average retail price of $2.13. The population standard deviation is known to be $0.20.

A. What do you need to assume in order to compute a confidence interval for the interval for the population mean retail price charged across the country.

B. Compute a 99% confidence interval for the population mean retail price. Interpret your answer in terms of the question.

C. Suppose you wish to estimate the population mean retail price to within $0.05 with 99% confidence interval. Retailers should be sampled in order to achieve the desired margin of error?

Answer 1

Answer:

a)

Given,

Here in the event that x1,x2,.......xn is an irregular example from an obscure probability distribution with mean and unknown variance then for huge n, we accept z = (xbar - mean)/(standard deviation/n) pursues N(0,1) roughly because of control limit theorem.

The test can be utilized as common in the circumstance.

b)

To give the 99% confidence interval

Interval = (xbar - z*/sqrt(n) , xbar + z*/sqrt(n) ) ----------> (1)

Here standard deviation is known

xbar = 2.13

z for 95% confidence interval is 2.58

standard deviation = 0.20

n = 25

Now substitute all values in equation(1)

Interval = (2.13 - 2.58*0.20/sqrt(25) , 2.13 + 2.58*.20/sqrt(25))

= (2.13 +/- 0.1032)

= (2.13 - 0.1032 , 2.13 + 0.1032)

= (2.0268 , 2.2332)

= (2.03 , 2.23)

So we can clearly say that the population mean retail price lies between $2.03 & $2.23