Question

An advertisement for a new toothpaste claims that it reduces cavities of children in their cavity-prone years. Cavities per year for their age group are normal with mean 3 and standard deviation 1.
A study of 2500 children who used their toothpaste 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.
(a) Are these data strong enough, at the 5 percent level of significance, to establish the claim of the toothpaste advertisement? Use the test statistic approach.
(b) Without doing any recalculation, do you think there would be a change in the conclusion in part (a) if you decrease the level of significance to 1 percent? Explain your answer.

Answer 1

Answer:

strong enough, at the 5 percent level of significance, to establish the claim of the toothpaste advertisement

Step-by-step explanation:

Let X be the Cavities per year for their age group

X is N(3,1)

Population mean and std deviation are given here

Sample size n = 2500

Sample mean = 2.95

Let us create hypotheses as

$H_0: \bar x = 3\\H_a: \bar x$

(left tailed test)

Mean difference = $2.95-3=-0.05$

Std error of mean = $\frac{\sigma}{\sqrt{n} } \\=0.02$

Z= mean difference/std error = -2.5

Level of significance =5%

For one tailed test at 5% significance level Z critical = -1.645

Since -2.5 <-1.645 we reject null hypothesis

These data are  strong enough, at the 5 percent level of significance, to establish the claim of the toothpaste advertisement