Question

A researcher is examining preferences among four new flavors of ice cream. A sample of n = 80 people is obtained. Each person tastes all four flavors and then picks a favorite. The distribution of preferences is as follows. Do these data indicate any significance preferences among the four flavors? Test at the .05 level of significance.

Ice Cream Flavor

A 12

B 18

C 28

D 22

Answer 1

Answer on Question #61378 – Math – Statistics and Probability

Question

A researcher is examining preferences among four new flavors of ice-cream. A sample of

$n = 80$ people is obtained. Each person tastes all four flavors and then picks a favorite. The distribution of preferences is as follows:

Do these data indicate any significance preferences among the four flavors? Test at the 0.05 level of significance.

Solution

We shall use the Chi-square test. Let

$H_0$ : the numbers of people preferring the new flavors of ice-cream are equal for all the four types;

$H_{1}$ : there is significance difference between the numbers of people preferring the new flavors of ice-cream.

The expected number of people preferring the one type of flavor (assuming that the $H_0$ is true) is equal to $\frac{n}{4} = \frac{80}{4} = 20$ .

Now we calculate the next value of test statistics:

$$\chi^ {2} = \frac {(1 2 - 2 0) ^ {2}}{2 0} + \frac {(1 8 - 2 0) ^ {2}}{2 0} + \frac {(2 8 - 2 0) ^ {2}}{2 0} + \frac {(2 2 - 2 0) ^ {2}}{2 0} = 6. 8.$$

Using the table of Chi-square distribution we find the critical value

$$\chi^ {2} (k = 4 - 1, \alpha = 0. 0 5) = \chi^ {2} (k = 3, \alpha = 0. 0 5) = 7. 8.$$

Since $6.8 < 7.8$ we have no reasons to reject $H_0$ and we conclude that these data do not indicate any significance preferences among the four flavors at the 0.05 level of significance.

Answer. These data do not indicate any significance preferences among the four flavors.

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