In a previous poll, 37% of adults with children under the age of 18 reported that their family ate dinner together seven nights a week. Suppose that, in a more recent poll, 396 of 1132 adults with children under the age of 18 reported that their family ate dinnertogether seven nights a week. Is there sufficient evidence that the proportion of families with children under the age of 18 who eat dinner together seven nights a week has decreased? Use the a= 0.05 significance level.Because npo (1-pPo) = 10 and the sample size is(Round to one decimal place as needed.)5% of the population size, and the sampleV the requirements for testing the hypothesisV satisfiedWhat are the null and alternative hypotheses?V versus H, OHo(Type integers or decimals. Do not round.)Find the test statistic, ZoZ0= (Round to two decimal places as needed.)Find the P-valueP-value = (Round to three decimal places as needed.)Is there sufficient evidence that the proportion of families with children under the age of 18 who eat dinner together seven nights a week has decreased? Choose the correct answer belowO A. Yes, there is sufficient evidence because the P-value is greater than the level of significance. Therefore, reject the null hypothesis.OB. No, there is not sufficient evidence because the P-value is greater than the level of significance. Therefore, reject the null hypothesis.Oc. Yes, there is sufficient evidence because the P-value is greater than the level of significance. Therefore, do not reject the null hypothesis.OD. No, there is not sufficient evidence because the P-value is greater than the level of significance. Therefore, do not reject the null hypothesis.

Given data number of success , x= 396 sample size, n =1132 population proportion,p= 0.37 Significance level, α= 0.05solution : a) npo(1-po) = 1132*0.37(1-0.37 ) = 263.9 because npo(1-po) = 263.9 >= 10 and the sample size is less than 5% of the population size, and the sample can be reasonably assumed to be random, the requirements for testing the hypothesis are satisfied. b) Hypothesis test : The null and alternative hypothesis is Ho : p = 0.37 H1 : p < 0.37 The hypothesis test is a left tailed testc) Test statistic : Test statistic for one proportion z test is =0.349823 -0.37 /√(0.37 (1-0.37 )/1132 = -1.41 The test statistic is -1.41d) P-value : p-value with test statistic z=-1.41is P-value =P(Z < Zobs ) = P(z <-1.41 ) =0.079 Excel formula: =NORM.DIST(-1.41,0,1,TRUE) ∴ P-value is 0.079e) Decision : since P-value is greater than significance level,α = 0.05 ,we failed to reject the null hypothesis Decision is fail to reject Ho therefore option d) is correct answer No, there is not sufficient evidence because the P-value is greater than the level of significance Therefore, do not reject the null hypotheses