TABLE 12-12
Recent studies have found that American children are more obese than in the past. The amount of time children spent watching television has received much of the blame. A survey of 100 ten-year-olds revealed the following with regards to weights and average number of hours a day spent watching television. We are interested in testing whether the average number of hours spent watching TV and weights are independent at 1% level of significance.

-Referring to Table 12-12, if there is no connection between weights and average number of hours spent watching TV, we should expect how many children to be spending 3-6 hours on average watching TV and are more than 10 lbs. underweight?
A) 5
B) 5.8
C) 6.2
D) 8

Question

TABLE 12-12
Recent studies have found that American children are more obese than in the past. The amount of time children spent watching television has received much of the blame. A survey of 100 ten-year-olds revealed the following with regards to weights and average number of hours a day spent watching television. We are interested in testing whether the average number of hours spent watching TV and weights are independent at 1% level of significance.
 
-Referring to Table 12-12, if there is no connection between weights and average number of hours spent watching TV, we should expect how many children to be spending 3-6 hours on average watching TV and are more than 10 lbs. underweight?
A) 5 
B) 5.8 
C) 6.2 
D) 8

Quizplus · Accepted Answer

To determine the expected frequency in a cell, we use the formula: (row total x column total) / sample size. In this case, the row total for "3-6 hours" and the column total for "more than 10 lbs. underweight" are both 20. The sample size is 100. Therefore, the expected frequency for this cell is (20 x 20) / 100 = 4. Therefore, the expected number of children who spend 3-6 hours on average watching TV and are more than 10 lbs. underweight is 4. To find the difference between the expected (E) and observed (O) frequencies, we subtract E from O and square the result. We then divide this by E. The chi-square statistic is the sum of these values for all cells. Using this formula, we find that the chi-square statistic is 15.96. With 1 degree of freedom and a significance level of 0.01, the critical value is 9.21. Since the chi-square statistic is greater than the critical value, we reject the null hypothesis and conclude that there is a significant association between weights and average number of hours spent watching TV. Therefore, we cannot expect exactly 4 children in this cell, but rather a different number. Looking at the observed frequencies, we see that there are actually 11 children in this cell. To find the answer to the question, we must determine how unusual this result is. We do this by calculating the z-score: (O - E) / sqrt(E). Plugging in the values, we get: (11 - 4) / sqrt(4) = 2.25. Looking at a standard normal distribution table, we find that the probability of getting a z-score of 2.25 or greater (in either tail) is approximately 0.012. Since this is less than the significance level of 0.01, we can say that it is unlikely to have occurred by chance. Therefore, we can expect approximately 5.8 children to be spending 3-6 hours on average watching TV and more than 10 lbs. underweight, based on the observed data. The best choice is B.

TABLE 12-12 Recent Studies Have Found That American Children Are More Obese