Of 150 adults who tried a new peach-flavoured peppermint patty, 90 rated it excellent. Of 200 children sampled, 123 rated it excellent. Using the 0.10 level of significance, can we conclude that there is a significant difference in the proportion of adults and the proportion of children who rate the new flavour as excellent? State the decision rule, the value of the test statistic, and your decision. A) Reject if z 1.645 or 1.645 or 1.645 or 1.96 or 1.96 or

Question

Of 150 adults who tried a new peach-flavoured peppermint patty, 90 rated it excellent. Of 200 children sampled, 123 rated it excellent. Using the 0.10 level of significance, can we conclude that there is a significant difference in the proportion of adults and the proportion of children who rate the new flavour as excellent? State the decision rule, the value of the test statistic, and your decision. A) Reject if z > 1.645 or < -1.645, z = -0.28, difference exists B) Reject if z > 1.645 or < -1.645, z = -0.28, no difference. C) Reject if z > 1.645 or < -1.645, z = -1.28, difference exists. D) Reject if z > 1.96 or < -1.96, z = -0.66, no difference. E) Reject if z > 1.96 or < -1.96, z = -2.26, difference exists.

Quizplus · Accepted Answer

We can use a two-sample z-test for proportions to compare the proportion of adults and children who rated the new flavour as excellent. The null hypothesis is that there is no difference in the proportion of adults and children who rated it excellent, while the alternative hypothesis is that there is a significant difference.

H0: p1 = p2
Ha: p1 ≠ p2

where p1 is the proportion of adults who rated it excellent and p2 is the proportion of children who rated it excellent.

Using the given information, we can calculate the sample proportions and the pooled proportion:

p1 = 90/150 = 0.6
p2 = 123/200 = 0.615

pooled proportion: 
p = (90+123) / (150+200) = 0.611

We can then calculate the test statistic:

z = (p1 - p2) / sqrt(p*(1-p)*(1/n1 + 1/n2))

z = (0.6 - 0.615) / sqrt(0.611*(1-0.611)*(1/150 + 1/200))

z ≈ -0.281

Using a significance level of 0.10 and a two-tailed test, the critical z-value is ±1.645. Since the calculated z-value (-0.281) is not in the rejection region, we fail to reject the null hypothesis at the 0.10 level of significance. Therefore, we conclude that there is no significant difference in the proportion of adults and children who rated the new flavour as excellent.

The correct answer is B.

Of 150 Adults Who Tried a New Peach-Flavoured Peppermint Patty