Question

A school has found from past records that the average scores for boys is 75 in a sample of 45 boys, and the average scores for girls is 85 in a sample of 30 girls. The population standard deviations for boys' and girls' scores were 25 and 17, respectively. The 90% confidence interval for the difference of the population means is approximately (_____, −2.022). Assume that the distributions of the scores of men and women are normally distributed. Round your answer to two decimal places.
Confidence Level Corresponding zc Value
90% confidence zc=1.645
95% confidence zc=1.96
99% confidence zc=2.576

Answer 1

Answer:

The 90% confidence interval for the difference of the population means is approximately (-17.98, -2.02).

Step-by-step explanation:

Before building the confidence interval, we need to understand the central limit theorem and subtraction of 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 $s = \frac{\sigma}{\sqrt{n}}$.

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.

Boys:

Mean of 75, sample of 45, standard deviation of 25.

This means that $\mu_B = 75, s_B = \frac{25}{\sqrt{45}}$

Girls:

Mean of 85, sample of 30, standard deviation of 17.

This means that $\mu_G = 85, s_G = \frac{17}{\sqrt{30}}$

Distribution of the difference of mean grades of boys and girls:

$\mu = \mu_B - \mu_G = 75 - 85 = -10$

$s = \sqrt{s_B^2+s_G^2} = \sqrt{(\frac{25}{\sqrt{45}})^2+(\frac{17}{\sqrt{30}})^2} = 4.85$

Confidence interval:

As stated, the critical value is $z = 1.645$

The margin of error is of:

$M = zs = 1.645*4.85 = 7.98$

Lower bound:

$\mu - M = -10 - 7.98 = -17.98$

Upper bound:

$\mu + M = -10 + 7.98 = -2.02$

The 90% confidence interval for the difference of the population means is approximately (-17.98, -2.02).