Question

A study of the effect of television commercials on 12-year-old children measured their attention span, in seconds. The commercials were for clothes, food, and toys.

Clothes Food Toys
43 30 52
24 38 58
42 46 43
35 54 49
28 47 63
31 42 53
17 34 48
31 43 58

Required:
a. Find the values of mean and standard deviation.
b. Is there a difference in the mean attention span Of the children for the various commercials?

Answer 1

Answer:

(a)

Mean

$\bar x_1 = 31.375$

$\bar x_2 = 41.75$

$\bar x_3 = 53.00$

Standard deviation

$\sigma_1 = 8.73$

$\sigma_2 = 7.65$

$\sigma_3 = 6.04$

(b) Yes, there is a difference in the mean

Step-by-step explanation:

Solving (a): The mean and standard deviation of each commercial

This is calculated as:

$\bar x = \frac{\sum x}{n}$

For clothes:

$\bar x_1 = \frac{43+24+42+35+28+31+17+31}{8}$

$\bar x_1 = \frac{251}{8}$

$\bar x_1 = 31.375$

For food:

$\bar x_2 = \frac{30+38+46+54+47+42+34+43}{8}$

$\bar x_2 = \frac{334}{8}$

$\bar x_2 = 41.75$

For toys:

$\bar x_3 = \frac{52+58+43+49+63+53+48+58}{8}$

$\bar x_3 = \frac{424}{8}$

$\bar x_3 = 53.00$

The sample standard deviation is:

$\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{n-1}}$

For clothes:

$\sigma_1 = \sqrt{\frac{(43 - 31.375)^2 +.............+(31 - 31.375)^2}{8-1}}$

$\sigma_1 = \sqrt{\frac{533.875}{7}$

$\sigma_1 = \sqrt{76.2678571429}$

$\sigma_1 = 8.73$

For food:

$\sigma_2 = \sqrt{\frac{(30 - 41.75)^2 +............+(43 - 41.75)^2}{8-1}}$

$\sigma_2 = \sqrt{\frac{409.5}{7}}$

$\sigma_2 = \sqrt{58.5}$

$\sigma_2 = 7.65$

For toys:

$\sigma_3 = \sqrt{\frac{(52-53.00)^2+...................+(58-53.00)^2}{8}}$

$\sigma_3 = \sqrt{\frac{292}{8}}$

$\sigma_3 = \sqrt{36.5}$

$\sigma_3 = 6.04$

Solving (b): Difference in mean in the commercials;

In (a), we have:

$\bar x_1 = 31.375$

$\bar x_2 = 41.75$

$\bar x_3 = 53.00$

$\bar x_1 \ne \bar x_2 \ne \bar x_3$

Hence, there is a difference in their means