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:

The test statistic is [tex]z = -0.28[/tex]

Step-by-step 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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

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 [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]

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:

[tex]p_1 = \frac{16}{40} = 0.4, s_1 = \sqrt{\frac{0.4*0.6}{40}} = 0.0775[/tex]

13 of 30 at the other had this infection.

This means that:

[tex]p_2 = \frac{13}{30} = 0.4333, s_2 = \sqrt{\frac{0.4333*0.5667}{30}} = 0.0905[/tex]

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:

[tex]H_0: p_1 - p_2 = 0[/tex]

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

[tex]H_a: p_1 - p_2 \neq 0[/tex]

The test statistic is:

[tex]z = \frac{X - \mu}{s}[/tex]

In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis and s is the standard error

0 is tested at the null hypothesis:

This means that [tex]\mu = 0[/tex]

From the two samples:

[tex]p = p_1 - p_2 = 0.4 - 0.4333 = -0.0333[/tex]

[tex]s = \sqrt{s_1^2+s_2^2} = \sqrt{0.0775^2+0.0905^2} = 0.1191[/tex]

Value of the test statistic:

[tex]z = \frac{X - \mu}{s}[/tex]

[tex]z = \frac{-0.0333 - 0}{0.1191}[/tex]

[tex]z = -0.28[/tex]

The test statistic is [tex]z = -0.28[/tex]

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