One of your peers claims that boys do better in math classes than girls. Together you run two independent simple random samples

Question

3 months ago

8

One of your peers claims that boys do better in math classes than girls. Together you run two independent simple random samples

and calculate the given summary statistics of the boys and the girls for comparable math classes. In Calculus, 15 boys had a mean percentage of 82.3 with standard deviation of 5.6 while 12 girls had a mean percentage of 81.2 with standard deviation of 6.7. What assumptions need to be made in order to determine the 90% confidence interval for the difference in the mean percentage scores for the boys in calculus and the girls in calculus? Supposing the assumption is true, calculate the interval.

Mathematics

1 answer:

babunello [11.8K] · Answer 1 · 2026-03-04T03:20:56+03:00

Answer:

Step-by-step explanation:

Hello!

To determine whether boys excel in math classes compared to girls, two random samples were collected:

Sample 1

X₁: score achieved by a boy in calculus

n₁= 15

X[bar]₁= 82.3%

S₁= 5.6%

Sample 2

X₂: score obtained by a girl in calculus

n₂= 12

X[bar]₂= 81.2%

S₂= 6.7%

To estimate a confidence interval for the difference between the average percentages of boys and girls in calculus, it's essential that both variables come from normally distributed populations.

For utilizing a pooled variance t-test, it is also required that the population variances, though unknown, are assumed to be equal.

The confidence interval can then be calculated with:

[(X[bar]_1 - X[bar]₂) ± $t_{n_1+n_2-2;1-\alpha /2}$ * $Sa*\sqrt{\frac{1}{n_1} +\frac{1}{n_2} }$ ]

$Sa= \sqrt{\frac{(n_1-1)S^2_1+(n_2-1)S^2_2}{n_1+n_2-2} } = \sqrt{\frac{14*(5.6^2)+11*(6.7^2)}{15+12-2} }= 6.108= 6.11$

$t_{n_1+n_2-2;1-\alpha /2}= t_{15+12-2;1-0.05}= t_{25;0.95}= 1.708$

[(82.3 - 81.2) ± 1.708 * (6.11 * $\sqrt{\frac{1}{15}+\frac{1}{12} }$ ]

[-2.94; 5.14]

Using a 90% confidence level, the interval [-2.94; 5.14] is expected to encompass the true difference between the average percentages achieved by boys and girls in calculus.

I hope this is of assistance!