Problem 2 A smalltown newspaper recently reported that for f

Problem 2)

A small-town newspaper recently reported that for families residing in its circulation area, the distribution of the

weekly (per capita) expenditure for food consumed away from home has an average of $3.28 and a standard deviation of $1.12. In order to check this claim, an economist randomly samples 100 families residing in the area and monitors their expenditures for food consumed away from home for a week.

a. What is the mean of the sampling distribution of the mean weekly expenditure for food purchased away from home for a random sample of 100 families residing in the area?

b. What is the standard deviation of the sampling distribution of the mean weekly expenditure for food purchased away from home for a random sample of 100 families residing in the area?

c. What is the probability that the sample mean weekly expenditure for food purchased away from home will be at least $3.60?

d. In a sample of 100 families, the average weekly expenditure for food purchased away from home is $3.00. Find a 90% confidence interval for the mean weekly expenditure for food purchased away from home.

Solution

Let X be the random variable that weekly (per capita) expenditure for food consumed away from home.

X has mean (average) = $3.28

standard deviation (sd) = $1.12

In order to check this claim, an economist randomly samples 100 families residing in the area and monitors their expenditures for food consumed away from home for a week.

a. What is the mean of the sampling distribution of the mean weekly expenditure for food purchased away from home for a random sample of 100 families residing in the area?

mean of the sampling distribution of the mean weekly expenditure = (µ) = $3.28

b. What is the standard deviation of the sampling distribution of the mean weekly expenditure for food purchased away from home for a random sample of 100 families residing in the area?

SD = sd / sqrt(n)

where n is random sample of families residing in the area = 100

SD = 1.12 / sqrt(100) = 0.112

c. What is the probability that the sample mean weekly expenditure for food purchased away from home will be at least $3.60?

that is here we want to find the P(Xbar >3.60)

Convert Xbar into standard normal score.

z = (Xbar - µ) / [ sd/sqrt(n) ]

z = (3.60 - 3.28) / 0.112 = 2.8571

Now we have to find P(Z > 2.8571)

P(Z > 2.8571) = 1 - P(Z <=2.8571)

Because table doesn\'t gives right tail area it gives always left tail area.

This probability we can find by using EXCEL.

syntax :

=NORMSDIST(z)

where z = 2.8571

P(Z<=2.8571) = 0.997863

P(Z > 2.8571) = 1 - 0.997863 = 0.0021

d. In a sample of 100 families, the average weekly expenditure for food purchased away from home is $3.00. Find a 90% confidence interval for the mean weekly expenditure for food purchased away from home.

Xbar = $3.00

c = confidence level = 0.90

90% confidence interval for the mean weekly expenditure for food purchased away from home is,

Xbar - E < µ < Xbar + E

where E is margin of error.

E = Zc * / sqrt(n)

c = confidence level = 0.90

a = 1 - c = 1 - 0.9 = 0.1

a/2 = 0.1/2 = 0.05

(sd) = 1.12

n = 100

Zc we can find by using EXCEL.

syntax :

=NORMSINV(probability)

probability = 1 - a/2 = 1-0.05 = 0.95

Zc = 1.645

E = 1.645 * 1.12/sqrt(100)

E = 1.645 * 0.112 = 0.1842

lower limit = Xbar - E = 3.00 - 0.1842 = 2.8158

upper limit = Xbar + E = 3.00 + 0.1842 = 3.1842

90% confidence interval for  the mean weekly expenditure for food purchased away from home is (2.8158, 3.1842).