The average number of times Americans dine out in a week fell from 4.0 in 2008 t

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 dines out last week provides the following data.

6             1              5              3              7              3              0              3              1              3

4              1              2              4              1              0              5              6              3              1

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

b) Do the data contains outliers

Explanation / Answer

(a)
Given:
Skewness = 0.34

This shows the curve is positively skewed,i,e., The mean is greater than the mode. The tail on the curve's right hand side is longer than the tail on the left hand side.

Is the shape you would expect?:

From the given data, mean = 59/20 = 2.95

SD = s = 2.0894

Q1 = First Quartile = 1

Q3 = Third Quartile = 4.75

We note that there many above the mean, like: 4,5,6,7. So, The positive skewness shape is expected.

(b)

To find outliers:

IQR = Q3 - Q1 = 4.75 - 1 = 3.75

So,

1.5 X IQR = 1.5 X 3.75 = 5.625

Q3 + 1.5 IQR = 4.75 + 5.625 = 10.375

It is seen that no values are > 10.375.

So, there are no outliers on high end of the distribution.

Q1 - 1.5 IQR = 1 - 5.625 = - 4.625.

It is seen that no values are < - 4.625.

So,

There are no outliers on the low end of the distribution.

Thus,

The data does not contain outliers.

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