Question

Many elementary school students in a school district currently have ear infections. A random sample of children in two different schools found that 16 of 40 at one school and 13 of 30 at the other had this infection. Conduct a test to answer if there is sufficient evidence to conclude that a difference exists between the proportion of students who have ear infections at one school and the other. Find the test statistic. [Suggestion: try to use a TI 83 or a similar calculator.]

Answer 1

The test statistic is
`z = -0.28`

Explanation:

First, before finding the test statistic, we need to understand the central limit theorem and difference between normal variables.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean
`\mu = p` and standard deviation
`s = \sqrt{(p(1-p))/(n)}`

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

A random sample of children in two different schools found that 16 of 40 at one school

This means that:

`p_1 = (16)/(40) = 0.4, s_1 = \sqrt{(0.4*0.6)/(40)} = 0.0775`

13 of 30 at the other had this infection.

This means that:

`p_2 = (13)/(30) = 0.4333, s_2 = \sqrt{(0.4333*0.5667)/(30)} = 0.0905`

Conduct a test to answer if there is sufficient evidence to conclude that a difference exists between the proportion of students who have ear infections at one school and the other.

At the null hypothesis, we test if there is no difference, that is:

`H_0: p_1 - p_2 = 0`

And at the alternate hypothesis, we test if there is difference, that is:

`H_a: p_1 - p_2 \\eq 0`

The test statistic is:

`z = (X - \mu)/(s)`

In which X is the sample mean,
`\mu` is the value tested at the null hypothesis and s is the standard error

0 is tested at the null hypothesis:

This means that
`\mu = 0`

From the two samples:

`p = p_1 - p_2 = 0.4 - 0.4333 = -0.0333`

`s = √(s_1^2+s_2^2) = √(0.0775^2+0.0905^2) = 0.1191`

Value of the test statistic:

`z = (X - \mu)/(s)`

`z = (-0.0333 - 0)/(0.1191)`

`z = -0.28`

The test statistic is
`z = -0.28`