Question

The table below gives beverage preferences for random samples of teens and adults.

	teens	adults	total
coffee	50	200	250
tea	100	150	250
soft drinks	200	200	400
other	50	50	100
	400	600	1000

a. We are asked to test for independence between age (i.e., adult and teen) and drink preferences. With a .05 level of significance, the critical value for the test is?

(a. 1.645 b. 7.815 c. 14.067 d. 15.507)

b. The expected number of adults who prefer coffee is? (a. 0.25 b. 0.33 c. 150 d. 200)

c. The test statistic for this test of independence is? (a. 0 b. 8.4 c. 62.5 d. 82.5)

d. The conclusion of the test is that the:

(Age is independent of drink preference, Age is not independent of drink preference, test is inconclusive, None of these alternatives is correct.)

Answer 1

Concepts and reason

Chi-square distribution:

Chi-square distribution with k degrees of freedom is defined as the sum of the k standard normal variables. This is also defined as the square of the deviations taken from the mean. Chi-square distribution is used for testing goodness of fit and independence in contingency tables.

Statistical hypotheses testing: Hypotheses testing is used to make inferences about the population based on the sample data. The hypotheses test consists of null hypothesis and alternative hypothesis.

For testing the mean, if the population standard deviation is known then the Z test is used. If the population standard deviation is not known then t test is used.

Null hypothesis: The null hypothesis states that there is no difference in the test, which is denoted by ${H_0}$ . Moreover, the sign of null hypothesis is equal $\left( = \right)$ , greater than or equal $\left( \ge \right)$ and less than or equal $\left( \le \right)$ .

Alternative hypothesis: The hypothesis that differs from the ${H_0}$ is called alternative hypothesis. This signifies that there is a significant difference in the test. The sign of alternative hypothesis is less than $\left( < \right)$ , greater than $\left( > \right)$ , or not equal $\left( \ne \right)$ .

Level of significance:

The level of significance is the probability of rejecting the null hypothesis given that it is true. This is also known as probability of type I error. It is denoted by $\alpha$ .

Fundamentals

The formula to find the degrees of freedom:

$df = \left( {r - 1} \right)\left( {c - 1} \right)$

Rejection rule based on critical value:

If ${\rm{Test statistic}} \ge {\chi ^2}_{{\rm{crit}}}$ , then reject the null hypothesis ${H_0}$ .

If ${\rm{Test statistic}} \le - {\chi ^2}_{{\rm{crit}}}$ , then reject the null hypothesis.

Excel add-in (MegaStat) procedure:

1.In EXCEL, Select Add-Ins > MegaStat > Chi-square/Crosstab > Contingency Table.

2.In Input range, select the cell range A1:A11.

3.In Output Options, select chi-square, Expected values, O –E, (O-E)²/E.

4.Click OK.

(a)

The critical value is obtained as shown below:

From the information, there are 4 rows and 2 columns, that is $r = 4,c = 2$ . The degrees of freedom is,

$\begin{array}{c}\\df = \left( {r - 1} \right)\left( {c - 1} \right)\\\\ = \left( {4 - 1} \right)\left( {2 - 1} \right)\\\\ = 3\\\end{array}$

Instructions to find the critical value $\chi _{{\rm{critical value}}}^2$ with the significance level 0.05 is,

In ${\chi ^2}$ distribution table, locate the 3^rd row corresponding to degrees of freedom. Now in 3^rd row locate the column where the level of significance is 0.05. The critical value for ${\chi ^2}$ distribution with 3 degrees of freedom for two tail test with the 0.05 level of significance is 7.81.

(b)

The expected number of adults who prefer coffee is obtained below:

Instructions to find the chi-square independent test:

1.In EXCEL, Select Add-Ins > MegaStat > Chi-square/Crosstab > Contingency Table.

2.In Input range, select the cell range A1:C5.

3.In Output Options, select chi-square, Expected values, (O-E)²/E.

4.Click OK.

Follow the above instructions to obtain following output,

From the output, the expected number of adults who prefer coffee is 150.

(c)

The test statistic is obtained below:

Follow the above instructions to obtain following output,

From the output, the test statistic is 62.50.

(d)

The null and alternative hypotheses are stated below:

Null hypothesis:

${H_0}:$ The Age is independent of drink preference.

Alternative hypothesis:

${H_a}:$ The Age is not independent of drink preference.

The conclusion is stated below:

The value of test statistic is 62.5 and the critical value is 7.815.

The test statistic is greater than the critical value.

That is, $\chi _{{\rm{data}}}^2\left( { = 62.5} \right) > \chi _{{\rm{critical}}}^2\left( { = 7.815} \right)$ .

By the rejection rule, reject null hypothesis at 0.05 level of significance.

Therefore, it can be concluded that the age is not independent of drink preference.

Ans: Part a

The critical value is 7.815.

Part b

The expected number of adults who prefer coffee is 150.

Part c

The test statistic is 62.5.

Part d

The age is not independent of drink preference.