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Results on seat belt usage from the 2003 Youth Risk Behavior Survey were published in a...

Results on seat belt usage from the 2003 Youth Risk Behavior Survey were published in a USA Snapshot on January 13, 2005. The following table outlines the results from the high school students who were surveyed in the state of Nebraska. They were asked whether or not they rarely or never wear seat belts when riding in someone else's car. Using α = .05, does this sample present sufficient evidence to reject the hypothesis that gender is independent of seat belt usage? Female Male Rarely or never use seat belt 212 324 Uses seat belt 1205 1162

(a) Find the test statistic. (Give your answer correct to two decimal places.)

(ii) Find the p-value. (Give your answer bounds exactly.) < p

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Answer #1

The following cross tablulation have been provided. The row and column total have been calculated and they are shown below Row 1 Row 2 Total Column 1 212 1205 1417 Column 2 324 1162 1486 Total 536 2367 2903 The expected values are computed in terms of row and column totals. In fact, the formula is Eり- T-, where Ri corresponds to the total sum of elements in row i Cj corresponds to the total sum of elements in column j, and T is the grand total The table below shows the calculations to obtain the table with expected values: Expected Values Column 1 Column2 Total 1417 536/2903 261.63 1483 536/2903- 274.37 Row 1 536 1417 2367/2903 1486 2367/2903 Row 2 2367 1155 37 1211.83 Total 1417 1486 2903 Based on the observed and expected values, the squared distances can be computed according to the following formula: (E -OE. The table with squared distances is shown below Expected Values Column 1 Column2 Total 1417 536/2903- 261.63 1483 536/2903- 274.37 Row 1 536 1417 2367/2903 1486 2367/2903 Row 2 2367 1155.37 1211.63 Total 1417 1486 2903 The following null and alternative hypotheses need to be tested: Ho: The two variables are independent Ha: The two variables are dependent This corrosponds to a ChiSquare test of independence. Based on the information provided, the significance level is a 0.05, the number of degrees of freedom is df (2 1 x (2 1), sothen the rejection region for this test is R-x:x2 3.841) The Chi Squared statistic is computed as follows: (Oy Eur -9.415 +2.132 +8.977 +2.033-22.557 3-1 Since it is observed that x222.5573.841, it is then concluded that the null hypothesis is rejected It is concluded that the nuil hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the two variables are dependent, at the 0.05 significance evel

A)

TS = 22.56

b)

p-value =

0.000002
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