5. An advertisement for a toothpaste claims that use of the product signif- ican
ID: 3325871 • Letter: 5
Question
5. An advertisement for a toothpaste claims that use of the product signif- icantly reduces the number of cavities of children in their cavity-prone years. Cavities per year for this age group are normal with mean 3 and standard deviation 1. A study of 2500 children who used this tooth- paste found an average of 2.95 cavities per child. Assume that the standard deviation of the number of cavities of a child using this new toothpaste remains equal to 1. 84 The The o o ae of ea glatin400 (a) Are these data strong enough, at the 5 percent level of signifn- cance, to establish the claim of the toothpaste advertisement? (b) Is this a significant enough reason for your children to switch to this toothpaste?Explanation / Answer
Solution:-
a)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: > 3.0
Alternative hypothesis: < 3.0
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample z-test.
Analyze sample data. Using sample data, we compute the standard error (SE), z statistic test statistic (z).
SE = s / sqrt(n)
S.E = 0.02
z (x - ) / SE
z = - 2.5
where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.
The observed sample mean produced a z statistic test statistic of - 2.5.
Thus the P-value in this analysis is 0.0062.
Interpret results. Since the P-value (0.0062) is less than the significance level (0.05), we have to reject the null hypothesis.
b) Yes, There is significant evidence in the favor of the claim that use of product significantly reduces the number of cavities of children in their cavity prone years.
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