Question

The average number of times Americans dine out in a week fell from 4.0 in 2008 to 3.8 in 2012 (Zagat.com, April 1, 2012). The number of times a sample of 20 families dined out last week provides the following data: 6 1 5 3 7 3 0 3 1 3 1 2 4 1 0 5 6 3 1 4

Complete the following in Excel:

a. Compute the mean and median.

b. Compute the first and third quartiles.

c. Compute the range and interquartile range.

d. Compute the variance and standard deviation.

e. The skewness measure for these data is 0.34. Comment on the shape of this distribution. Is it the shape you would expect? Why or why not?

f. Do the data contain outliers?

Answer 1

The data is entered in the cells A1 to A20.

Excel formulae are given within the brackets for the following:

a.

Mean =2.95 (=AVERAGE(A1:A20))

Median =3 (=MEDIAN(A1:A20))

b.

First quartile, Q1 =1 (=QUARTILE(A1:A20, 1))

Third quartile, Q3 =4.25 (=QUARTILE(A1:A20, 3))

c.

Range =7 (=A25 - A26)

Interquartile Range, IQR=3.25 (=A24 - A23)

To find range:

Maximum value =7 (=MAX(A1:A20))

Minimum value =0 (=MIN(A1:A20))

d.

Variance =4.3658 (=VAR(A1:A20))

Standard deviation =2.0894 (=STDEV(A1:A20))

e.

Since the skewness of 0.34 is between -0.5 and 0.5, the shape of this distribution is approximately symmetrical (normal).

Yes. It is the shape I would expect because more number of families would be clustered around the center and it decreases as we move towards left and right from the center making the distribution to be normal or approximately normal.

f.

1.5(IQR) =1.5(3.25) =4.875

Q1 - 1.5(IQR) =1 - 4.875 = -3.875

Q3 + 1.5(IQR) =4.25 + 4.875 =9.125

Since, the data do not contain any values that are less than -3.875 or greater than 9.125, the data do not contain outliers.