4 Recall that the researchers reported the mean and the standard deviation of the anxiety scale forthe sample (M = 33.56, SD = 19.40) as well as for the population (µ = 14.20, o = 9.70), and that thesample size was N= 92. As mentioned on a previous screen, the researchers reported, “A singlesample z test revealed that the average score on the SCAS-P for the sample of children with ASsignificantly differed from the mean of the normative sample, z = 19.15, p< .05" (pp. 364–365). Oneassumption when performing a z test is that scores in the underlying population from which thesample is drawn are approximately normally distributed. These researchers did not report any testsof this assumption. Let's assume the worst-case scenario-that the anxiety scores in the populationare not normally distributed. What should we conclude about this z test? (Hint: Think of the central limit theorem.)O A. The results of the z test are not at all valid because the researchers failed to meet the assumption ofhaving normally distributed scores in the population.O B. The central limit theorem tells us that the distribution of sample means will approach normalitywhen the sample size is greater than 30, even if the underlying distribution of scores in thepopulation is non-normal. Given that the sample size in this study is 92, the researchers are okaywith using the z test.O C. We will need to penalize the p value for the z test because the data failed to meet the normalityassumption. The new, penalized p value is likely to be larger than 0.05.O D. The central limit theorem tells us that the distribution of sample means will approach normalitywhen the sample size is greater than 80, even if the underlying distribution of scores in thepopulation is non-normal. Given that the sample size in this study is 92, the researchers are okaywith using the z test.

The question is about central limit theorem Given : The popl mean ( μ ) = 14.2 The popl. std…

Recall that the researchers reported the mean and the standard deviation of the anxiety scale for the sample (M= 33.56, SD = 19.40) as well as for the population (µ = 14.20, o = 9.70), and that the sample size was N= 92. As mentioned on a previous screen, the researchers reported, “A single sample z test revealed that the average score on the SCAS-P for the sample of children with AS significantly differed from the mean of the normative sample, z= 19.15, p< .05" (pp. 364–365). One assumption when performing a z test is that scores in the underlying population from which the sample is drawn are approximately normally distributed. These researchers did not report any tests of this assumption. Let's assume the worst-case scenario-that the anxiety scores in the population are not normally distributed.