Question

I'm not sure how to enter the data. I've tried and it looks wrong. Also C, D, E, F is what I need help with as well. I've attached a copy of my output please tell me what I am doing wrong. I would like to know how to enter the data correctly.

Out put I keep getting but it looks as if it's off. what am I doing wrong?

Exercise for Chapter 17 You will be analyzing the data in Table 17.3 on the next page. Twenty high school math teachers were surveyed to determine whether they approved of a proposed new mathematics cur- riculum. The 20 teachers were classified as either experienced or inexperienced. Conduct a chi- square test of independence to test between Experience and Approval only the raw numbers in each category but also the associated percentages. This is im- i ortant when the groups are not equal in size Ihe cross tabulation table in Example 17.2 (based on the SPSS Statistics output in Figure 17.11 on the previous ge has been reformatted to be consistent with the style suggested in the Publication Manual of the American Psy- ological Association In the Variable View" mode, name the first variable Experience, and use the value l bels of 1- Experienced and 2-Inexperienced. Name the second variable Approval, and value labels of 1 Approve and 2 Disapprove. Following the previous example in Step 11 on page 180, move Experience to the "Row(s)" box. As in Step 12, move Approval to the "Column(s)" box. Table 17.3 Results of a Survey of Teachers on a Proposed New Mathematics Curriculum and Teachers Experience in Teaching Voter Number Teaching Experience Codes for Approve of Codes for Curriculum? Approval Experience Experienced Approve Approve Experienced Experienced Experienced Experienced Experienced Experienced Experienced Experienced Experienced Inexperienced Inexperienced Inexperienced Inexperienced Inexperienced Inexperienced Inexperienced Inexperienced Inexperienced Inexperienced Approve Approve Approve Approve Disapprove Disapprove Disapprove Approve Approve Approve 10 12 13 Disapprove Disapprove Disapprove Disapprove 15 17 19 Disapprove a. How many of the experienced teachers approved? b. How many of the inexperienced teachers approved? c. What is the value of chi-square? d. What is the associated probability? Are the results statistically significant at the 05 level? f. Write a statement of the results of the significance test. Crosstabs Case Processing Summary Cases Missing Total Valid Percent Percent N Percent 19 100.0% 19 100.0% 0.0% 0 Experience Approval Experience Approval Crosstabulation Approval Total 10 Count Experience 1 70.0% 30.0% 100.0% %within Experience 2 Count 33.3% 10 52.6% 66.7% 100.0% 19 %within Experience Count %within Experience Total i00.0% 47.4% Chi-Square Tests Asymptotic ance (2-sided) Exact Sig. (1-sided) Exact Sig (2-sided) df Value 110 255 106 2.554 Pearson Chi-Square Contnuity Correction 1.295 2.612 Likelihood Ratio Fisher's Exact Test Linear-by-Linear 128 179 120 2.420 Association N of Valid Cases 19 a. 3 cells (75.0%) have expected count less than S. The minimum expected count is 4.26. b. Computed only for a 2x2 table MacBook Ai

Answer 1

This is a simple problem related to development of a contigency table and then to conduct a chi square test to check for the significance.

So let us develop the contigency table first .

Please note that we have taken the rows as experienced /inexperienced and columns as approve and disapprove respectively.

	Approve	Disapprove	Total
Experienced	7	3	10
Inexperienced	3	7	10
Total	10	10	20

The following cross tablulation have been provided. The row and column total have been calculated and they are shown below: Total 10 10 20 Column 2 Column 1 Row 1 Row 2 Total 10 10 The expected values are computed in terms of row and column totals. In fact, the formula is Eij-TG, where R, corresponds to the total sum of elements in row i, Ci 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 Total Column 2 10x15 1015 Column 1 1o105 1o105 10 Expected Values 10 Row 1 20 10 Row 2 20 20 10 Total Based on the observed and expected values, the squared distances can be computed according to the following formula: (E O)2/E. The table with squared distances is shown below: Column2 Column 1 3-5E0.8 50.8 Squared Distances Row 1 Row 2 50.8 = 0.8

The following null and alternative hypotheses need to be tested: HO: The two variables are independent H The two variables are dependent This corresponds to a Chi-Square test of independence. Reiection Recion 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) 1, so then the rejection region for this test is R-[x: x2> 3.841) The Chi-Squared statistic is computed as follows: x®S(Qj E E,,)2=0.8 +0.8 +0.8 +0.8 = 3.2 Since it is observed that x-3.2 Sx rejected. 3.841, it is then concluded that the null hypothesis is not It is concluded that the null hypothesis Ho is not rejected. Therefore, there is NOT enough evidence to claim that the two variables are dependent, at the 0.05 significance level.

Thus we conclude that the approval rating is NOT dependent on the experience of the participants. (Both variables are NOT related to each other)

Other answers from the contigency table.

1. 7 experienced teachers approved.

2. 3 inexperienced teachers approved.

3. The value of chi square is 3.20

4. The associated probability for chi square value of 3.2 is 0.0736

5. The test signifies that the null hypothesis cant be rejected meaning that the two variables are NOT dpendent.

6. We conclude that the approval rating is NOT dependent on the experience of the participants. (Both variables are NOT related to each other)