Mathematics, 13.07.2021 23:30 zanedog2018

Would you favor spending more federal tax money on the arts? Of a random sample of n1 = 222 women, r1 = 51 responded yes. Another random sample of n2 = 174 men showed that r2 = 49 responded yes. Does this information indicate a difference (either way) between the population proportion of women and the population proportion of men who favor spending more federal tax dollars on the arts? Use = 0.05.

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The p-value of the test is 0.242 > 0.05, which means that this information does not indicate a difference between the population proportion of women and the population proportion of men who favor spending more federal tax dollars on the arts.

Step-by-step explanation:

Before solving this question, we need to understand the central limit theorem and subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem establishes 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 $s = \frac{\sigma}{\sqrt{n}}$ .

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 $s = \sqrt{\frac{p(1-p)}{n}}$

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.

Women:

51 out of 222, so:

$p_1 = \frac{51}{222} = 0.2297$

$s_1 = \sqrt{\frac{0.2297*0.7703}{222}} = 0.0282$

Men:

49 out of 174, so:

$p_2 = \frac{49}{174} = 0.2816$

$s_2 = \sqrt{\frac{0.2816*0.7184}{174}} = 0.0341$

Does this information indicate a difference (either way) between the population proportion of women and the population proportion of men who favor spending more federal tax dollars on the arts?

Either way, so a two tailed test to see if the difference of proportions is different of 0.

At the null hypothesis, we test if it is not different of 0, so:

H_0: p_1 - p_2 = 0

At the alternative hypothesis, we test if it is different of 0, so:

$H_1: p_1 - p_2 \neq 0$

The test statistic is:

$z = \frac{X - \mu}{s}$

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 samples:

X = p_1 - p_2 = 0.2297 - 0.2816 = -0.0519

$s = \sqrt{s_1^2+s_2^2} = \sqrt{0.0282^2+0.0341^2} = 0.0442$

Value of the test statistic:

$z = \frac{X - \mu}{s}$

$z = \frac{-0.0519 - 0}{0.0442}$

z = -1.17

P-value of the test and decision:

The p-value of the test is the probability of the differences being of at least 0.0519, either way, which is P(|z| > 1.17), that is, 2 multiplied by the p-value of z = -1.17.

Looking at the z-table, z = -1.17 has a p-value of 0.121.

0.121*2 = 0.242

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