Question

8:48 @ F . Rochelle Johnson-... 8:46 PM . • A random sample of students is interviewed to determine if between class and attitude toward · corporations. With the following results, test the hypothesis · that there is no difference among students on this attitude. · Favorable Neutral Unfavorable · Freshmen 100 50 70 · Sophomores 80 60 70 · Juniors 50 50 80 · Seniors 40 60 90 Show quoted text

Question 14.10


To answer this question you will have to follow the hypothesis testing steps provided in Chapter 14.

You will use the Chi-square bedasue the responses are classified into nominal categories.
Step 3. Significance level -Alpha for this study will be .05
You will now calculate observed and expected values as follows, I have added the first row of calculations and you can finish the rest.
Total % of Total Favorable
Interviewed Expected Freq
Freshmen 220 0.275 74.25
Soph 210
Junior 180
Senior 190
Totals 800


Step 4. Calculated Value
You will have to apply the formula to each of the three categories Favorable Neutral and Unfavorable for all 4 student classifications.
Begin with Fresmen Favorable


Apply the formula start with Freshmen favorable and go down the line until you have added all of them. Here is the first one as a sample:
8.93013468


You will also need to figure out the degrees of freedom. You have a four by four table so (4-1)+(4-1)=6


Step 5. Locate the .05 alpha & 6 d.f. on the table in Appendix D. This will gie you the crital test value.

Answer 1

solutely hypothesis are The null Ho: and alternative Mi = 2 Step 3: Total 220 7 freshnan Sophondes Tjunion 60 210 180 190 60 serons 800 220 310 to the stop a : Expected frequency favorable & Row to tu x Grand Colum total totas neutral unfavorable 85-25 freshonen 74.25 60.5 Sophorners 70-87 57.75 81.37 Junion 60.75 49.50 69.75 52.25 64.12 | senios 73.62 220 x 270/800 220 x 220/800 = 60.5

chi-square test statistic gnCo:-@)e/EL o observed 76.25 va (100–70.25)2, (50-60.5) ..t 10.5 160-5225) (90-73.62)? B. 62 Y : 8.93 +1.82 € 6-73 + 6.17 +0.09+ 1.59+1.90 +0.01 +1.51€ 9.08 +1.25 +3.64 = 33.61 steps! Significance level 42065 - Co-DxCc-1) = (4-1) x(3-1) = 342 = 6 Critical value of chi-square 2 = 12.592 since test statistic yz (33.61) is greater than criticar value of the ( 12.592) [33,613 12.591] so we reject nur hypothesis Ho. Hence it can be conclude that there is deperence among students on his attitude. nestl