The relative effectiveness of two different sales techniques was tested on 12 volunteer housewives.Each housewife was exposed to each sales technique and asked to buy a certain product, the sameproduct in all cases. At the end of each exposure, each housewife rated the technique with a 1 if she feltshe would have agreed to buy the product and a 0 if she probably would not have bought the product.Tech121 2 3 4 5Housewife0 161 1 1 1 1 01 0 0 07 8 90 0 100110Use large sample approximations and compare results.1011120 101a.Use Cochran's testb. Rearrange the data and use McNemar's testC.Ignore the blocking effect in this experiment and treat the data as if 24 difference housewiveswere used.Test Ho: P1=p2 versus Ha: p₁*P2

Given Information: Consider the given dataset: Housewife Technique 1 Technique 2 1 1 1 2 1 1 3 1 1 4 1 0 5 1 0 6 0 0 7 0 0 8 0 0 9 1 1 10 1 0 11 0 0 12 1 1a. The null and alternative hypotheses are: H0:The proportion of success is equal for all groupsHa: The proportion of success is different for atleast one group. The level of significance is 0.05. The number of groups we have : k=2 The degree of freedom is. df=k-1 =2-1=1 The test statistics can be obtained as: Q=k-1k∑j=1kx2j-N2kN-∑i=1nx2i The table that shows the computation is shown below: Housewife Technique 1 Technique 2 Total 1 1 1 2 2 1 1 2 3 1 1 2 4 1 0 1 5 1 0 1 6 0 0 0 7 0 0 0 8 0 0 0 9 1 1 2 10 1 0 1 11 0 0 0 12 1 1 2 Total 8 5 13 The test statistics can be obtained as: Q=2-12×82+52-1322×13-22+22+...02+22 =12×89-16926-23 =93 =3 Hence, the test statistic is 3. The critical value can be obtained by using the excel function =CHIINV(). The screenshot is shown below: Decision criterion: Reject the null hypothesis if Q>χ2critOtherwise fail to reject the null hypothesis. Conclusion: Since the test statistic is not greater than the critical value. Thus, the null hypothesis does not get rejected at 5% level of significance. Hence, we can conclude that the proportion of success if equal for all groups. b. Rearrange the data: No Yes Total No 4 (a) 0 (b) 4 Yes 4 (c) 4 (d) 8 Total 8 4 12 The test statistic can be calculated as: χ2=b-c2b+c =0-420+4 =164 =4 Hence, the test statistic is, The critical value can be obtained by using the excel function =CHIINV(). The screenshot is shown below: Decision criterion: Reject the null hypothesis if χ2obs>χ2critOtherwise fail to reject the null hypothesis. Conclusion: Since the test statistic is greater than the critical value. Thus, the null hypothesis does get rejected at 5% level of significance. There is a significant association between the paired nominal datac. Ignore the blocking effect in this experiment and treat the data as if 24 different housewives were used. The null and alternative hypotheses are: H0:p1=p2Ha:p1≠p2 From the above data: Technique 1 x1=8 , n1=12 Technique 2 x2=5 , n2=12 The sample proportion can be obtained as: p^1=812=0.667p^2=512=0.417 The polled proportion can be obtained as: P=8+512+12=1324=0.542 The standard error can be calculated as: S.E=P1-p1n1+1n2 =0.203 The test statistic can be calculated as: z=p^1-p^2S.E =0.667-0.4170.203 =1.232 The two-tailed critical value at 5% level of significance will be 1.96. Decision criterion: Reject the null hypothesis if Zobs>Z2critOtherwise fail to reject the null hypothesis. Conclusion: Since the test statistic is not greater than the critical value. Thus, the null hypothesis does not get rejected at a 5% level of significance. There is not a significant difference between the proportions.