Question

A study that looked at beverage consumption used sample sizes that were much smaller than previous national surveys. One part of this study compared 20 children who were 7 to 10 years old with 5 who were 11 to 13. The younger children consumed an average of 8.2 oz of sweetened drinks per day while the older ones averaged 14.5 oz. The standard deviations were 10.9 oz and 8.2 oz respectively.Use younger children as population 1 (a) Do you think that it is reasonable assume that these data are Normally distributed? Explain why or why not. (Hint: Think about the 68-95-99.7 rule.) The 68-95-99 rule suggests that the distributions are not Normal. If they were, we would expect some of the children to drink negative amounts of sweetened drinks, which does not make sense be Norma according to the 68-95-99.7 rule. We can see that the groups have means and standard deviations, two characteristics of Normal distributions. The distributions appear (b) Using the methods in this section, test the null hypothesis that the two groups of children consume equal amounts of sweetened drinks versus the two-sided allternative. Report all details of the significance-testing procedure with your conclusion. (Round your answers to three decimal places.) SEn t= df = P-value four decimal places.) (c) Give a 95% confidence interval for the difference in means. (Round your answers (d) Do you think that the analyses performed in parts (b) and (c) are appropriate for these data? Explain why or why not. The t procedures are appropriate for these data because the distributions are Normal and the samples are large enough. OThe t procedures are questionable for these data because the distributions are not Normal and the samples are small. (e) The children in this study were all participants in an intervention study at a summer day camp at a particular university. To what extent do you think that these results apply to other groups of children? We could easily extend these results to other children, because this group follows a Normal distribution and has reasonable sample sizes. We would have to be very cautious about extending these results to other children, because this group is not an SRS and might not be random in any way.

Answer 1

a)

The 68-95-99.7 rule suggests that the distributions are not Normal. If they were, we would expect some of the children to drink negative amounts of sweetened drinks, which does not make sense

b)

The provided sample means are shown below: X1 8.2 Х, 14.5 Also, the provided sample standard deviations are: 10.9 S1 8.2 S2 and the sample sizes are ni 20 and n2 = 5 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: 2 На: Д1 # M2 This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

(2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the degrees of freedom 23. In fact, the degrees of freedom are computed as follows, assuming that the population are df variances are equal: Hence, it is found that the criticcal value for this two-tailed test is te 23 2.069, for a 0.05 and df The rejection region for this two-tailed test is R {t : \t| > 2.069} (3) Test Statistics Since it is assumed that the population variances are equal, the t-statistic is computed as follows: X2 (n1-1)s(ng-1)s% n1+ng-2 ni n2 8.2 14.5 -1.202 (20-1)10.92(5-1)8.2 20+5-2 20 +

(4) Decision about the null hypothesis Since it is observed that t= 1.202 < te = 2.069, it is then concluded that the null hypothesis is not rejected Using the P-value approach: The p-value is p = 0.2415, and since p 0.2415 2 0.05, it is concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to is different than 2, at the 0.05 significance level. claim that the population mean Confidence Interval The 95% confidence interval is -17.14 < l1 - f2< 4.54

c)

Based on the information provided, we assume that the population variances are equal, so then the number of degrees of freedom are df = ni n2 - 2 = 20 + 5 - 2 = 23. The critical value for a = 0.05 and df = 23 degrees of freedom is te = t1_a/2:n-1 = 2.069. The corresponding confidence interval is computed as shown below: Since the population variances are assumed to be equal, we need to compute the pooled standard deviation, as follows: - 1)s(n2-1)s (n1 пі + n, — 2 (20 - 1) x 10.92 (5 - 1) x 8.22 20 5 2 10.481

Since we assume that the population variances are equal, the standard error is computed as follows: 1 Sp se -- + n2 1 1 10.481 x 20 5 5.24 Now, we finally compute the confidence interval: X2 te x se, X1 - X2 te x se) (T CI - (8.2 14.5 -- 2.069 x 5.24,8.2 - 14.5+ 2.069 x 5.24) (-17.14, 4.54) Therefore, based on the data provided, the 95% confidence interval for the difference between the population means 2is 17.14 < /ui -l2 < 4.54, which indicates that we are 95% confident that the true difference between population means is contained by the interval (-17.14,4.54)

d) If the data were normal, they were appropriate. e) Other groups of children results may vary because these children were part of an intervention study.