Question

A food company is concerned about recent criticism of the sugar content of their children’s cereals. The data show the sugar content (as a percentage of weight) of several national brands of children’s and adults’ cereals.

Children’s cereals: 40.3, 55, 45.7, 43.3, 50.3, 45.9, 53.5, 43, 44.2, 44, 47.4, 44, 33.6, 55.1, 48.8, 50.4, 37.8, 60.3, 46.6

Adults’ cereals: 20, 30.2, 2.2, 7.5, 4.4, 22.2, 16.6, 14.5, 21.4, 3.3, 6.6, 7.8, 10.6, 16.2, 14.5, 4.1, 15.8, 4.1, 2.4, 3.5, 8.5, 10, 1, 4.4, 1.3, 8.1, 4.7, 18.4

a) Write the null and alternative hypotheses.

b) Check the conditions.

c) Find the 95% confidence interval for the difference in means.

d) Is there a significant difference in mean sugar content between these two types of cereals? Explain.

Show all work.

Answer 1

a) NULL HYPOTHESIS H0:

ALTERNATIVE HYPOTHESIS Ha:

b) Let us check the most important assumption for this test is homogeneity of variances means equalty of population variances.

HERE POPULATIONS VARIANCES ARE EQUAL SINCE LEVENES TEST P VALUE= 0.231

c) Pooled Variance
s^2p = ((df1)(s^21) + (df2)(s^22)) / (df1 + df2) = 2306.13 / 45 = 51.25

Standard Error
s(M1 - M2) = √((s^2p/n1) + (s^2p/n2)) = √((51.25/19) + (51.25/28)) = 2.13

Confidence Interval
μ1 - μ2 = (M1 - M2) ± ts(M1 - M2) = 36.6464 ± (2.01 * 2.13) = 36.6464 ± 4.285589

μ1 - μ2 = (M1 - M2) = 36.6464, 95% CI [32.3608, 40.9320].

You can be 95% confident that the difference between your two population means (μ1 - μ2) lies between 32.3608and 40.9320

d) t= xbar-ybar/sp*sqrt(1/n1+1/n2)

t= 46.80-10.1536/2.31

t= 17.223

p value=0.000

Since p value for t test is 0.000 which is significant (less than 0.05 level of significance) hence we can reject H0 and conclude that we have sufficient evidence to show that there is difference in means.

NOTE: I HAVE DONE STEP BY STEP ALSO BY USING SPSS. PLEASE COMMENT FOR ANY DOUBT. THANK YOU