Please number the response thank you In a study of marital satisfaction, a randomsample of 12 couples married for betweenfive and fifteen years was obtained. Eachcouple was interviewed by the researcher andrated on a Marital Cohesion (MC) scale.MC MC214 1969 817 4914 19615 22513 16910 10017 28911 12118 32414 19614 196Σ- 156 Σ(x2)=2142A. What is the mean of Marital Cohesion?1/2B. What is the standard deviation of Scores forhypothesis testing? (Round to the nearesthundredths)C. Is there a significant difference betweenthe sample mean of scores and thepopulation mean of 15? Carry out all 9 stepsof the hypothesis testing. Summary of the 9 Steps for Hypothesis Testing1. Null and alternative hypothesis.2. Two-tailed test.3. Select a significant level alpha.4. Determine T test OR Z test.5. Terit OR Zerit (Draw a graph)6. Calculate Tobt or Zobt.7. Check whether the observed value (obt) falls in the criticalregion.8. Reject or Fail to Reject Null Hypothesis.9. Therefore, there isa statisticallysignificantdifferencebetween the sample mean and the population mean ORthere is no significant difference between the two means.

Given: Sample size n = 12 Formula Used: Mean X = ∑Xin Sample standard deviation S = ∑Xi-X2n-1 Test-statistic t = X-μSnSolution: A). The Mean of marital Cohesion X = 14+9+7+.........................+1412 = 15612 The Mean of marital Cohesion X = 13B). The standard deviation S = 14-132+(9-13)2+......................+(14-13)212-1 = 11411 = 3.22C). Null hypothesis: H0 : μ = 15 Alternative hypothesis: Ha : μ ≠ 15 The test is two tailed. Since, the significance level is not given, we'll assume that the significance level is 0.05 α = 0.05 Since, the population standard deviation is not given, so we'll use t-test. T-test will be used here. Degree of freedom = n-1 = 12-1 = 11 Critical value of t at 0.05 significance level for 11 df and for two tailed test is: Tcrit = 2.20Test-statistic Tobt = X-μSn = 13-153.2212 = -2.1517 Tobt = 2.1517 The observed value (obt) did not fall in critical region. Fail to reject null hypothesis. There is not significance difference between the two means. Answer: A). The Mean of marital Cohesion X = 13 B). The standard deviation of scores for hypothesis testing is = 3.22 C). 1. Null hypothesis: H0 : μ = 15 Alternative hypothesis: Ha : μ ≠ 15 2. The test is two tailed. 3. Significance level α = 0.05 4. T-test 5. Tcrit = 2.20 6. Tobt = 2.1517 7. The observed value (obt) did not fall in critical region. 8. Fail to reject null hypothesis. 9. There is not significance difference between the two means.