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Chapter 11: Q.56 (page 730)

The cell in the table that contributes the most to the chi-square statistic is (a) Female, 50–50 chance.

(b) Male, 50–50 chance.

(c) Female, almost certain.

(d) Male, almost certain.

(e) All the cells contribute equally to the test statistic.

Short Answer

Expert verified

The correct option is (d) Male, almost certain.

Step by step solution

01

Given Information

The table is

02

Explanation

To determine the cell that contributes more to the test statistic that need to be computed based on the expected frequency.

The chi-square test formula is

χ2=∑Oi-Ei2Ei

The Expected frequency formula is given below as,

Ei=Row total×Column totalGrand total

The expected frequency for observations is calculated as follows:

E1=222×26254877

E1=119.5

03

Explanation

From the above table, the last column represents the each cell contribution to test statistic.

It can be seen that 20.51 is the largest value among them. It represents the value of Male, almost certain.

Thus, Male, almost certain cell contributes most to the test statistic.

Therefore correct option is (d) Male, almost certain.

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Most popular questions from this chapter

In the previous Check Your Understanding (page 698), we presented data on the use of Facebook by two randomly selected groups of Penn State students. Here are the data once again.

Do these data provide convincing evidence of a difference in the distributions of Facebook use among students in the two campus settings?

State appropriate null and alternative hypotheses for a significance test to help answer this question.

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(a) Should we use a chi-square test for homogeneity or a chi-square test of association/independence in this setting? Justify your answer. (b) State appropriate hypotheses for performing the type of test you chose in part (a). Minitab output from a chi-square test is shown below.

(c) Check that the conditions for carrying out the test are met.

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T11.5. We compute the value of theχ2statistic to be 6.58. Assuming that the conditions for inference are met, the P-value of our test is

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