Respuesta :
Answer:
Step-by-step explanation:
Hello!
To test if boys are better in math classes than girls two random samples were taken:
Sample 1
X₁: score of a boy in calculus
n₁= 15
X[bar]₁= 82.3%
S₁= 5.6%
Sample 2
X₂: Score in the calculus of a girl
n₂= 12
X[bar]₂= 81.2%
S₂= 6.7%
To estimate per CI the difference between the mean percentage that boys obtained in calculus and the mean percentage that girls obtained in calculus, you need that both variables of interest come from normal populations.
To be able to use a pooled variance t-test you have to also assume that the population variances, although unknown, are equal.
Then you can calculate the interval as:
[(X[bar]_1-X[bar_2) ± [tex]t_{n_1+n_2-2;1-\alpha /2}[/tex] * [tex]Sa*\sqrt{\frac{1}{n_1} +\frac{1}{n_2} }[/tex]]
[tex]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[/tex]
[tex]t_{n_1+n_2-2;1-\alpha /2}= t_{15+12-2;1-0.05}= t_{25;0.95}= 1.708[/tex]
[(82.3-81.2) ± 1.708* (6.11*[tex]\sqrt{\frac{1}{15}+\frac{1}{12} }[/tex]]
[-2.94; 5.14]
Using a 90% confidence level you'd expect the interval [-2.94; 5.14] to contain the true value of the difference between the average percentage obtained in calculus by boys and the average percentage obtained in calculus by girls.
I hope this helps!
The 90% confidence interval for the difference in the mean percentage scores for the boys in calculus and the girls in calculus is [-2.94 ; 5.14].
Given :
- One of your peers claims that boys do better in math classes than girls.
- 15 boys had a mean percentage of 82.3 with a standard deviation of 5.6.
- 12 girls had a mean percentage of 81.2 with a standard deviation of 6.7.
According to the given data:
For the first sample-
[tex]\rm X_1[/tex] : A boy score in calculus
[tex]\rm n_1 = 15\\[/tex]
[tex]\rm \overline{X_1}[/tex] = 82.3%
[tex]\rm S_1[/tex] = 5.6%
For the second sample-
[tex]\rm X_2[/tex] : A girl score in calculus
[tex]\rm n_2 = 12[/tex]
[tex]\rm \overline{X_2}[/tex] = 81.2%
[tex]\rm S_1[/tex] = 6.7%
Now, the interval can be calculated as:
[tex]\rm [\overline{X_1}-\overline{X_2}] \pm t_{n_1+n_2-2\;;\;1-\alpha /2}\times Sa\sqrt{\dfrac{1}{n_1}+\dfrac{1}{n_2}}[/tex]
[tex]\rm Sa=\dfrac{(n_1-1)S^2_1+(n_2-1)S^2_2}{n_1+n_2-2}[/tex]
[tex]\rm Sa = \dfrac{14\times 5.6^2+11\times 6.7^2}{15+12-2}[/tex]
Sa = 6.11
[tex]\rm t_{n_1+n_2-2\;;\;1-\alpha /2}= t_{15+12-2\;;\;1-0.05}= t_{25\;;\;0.95}=1.708[/tex]
[tex]\rm [82.3-81.2] \pm 1.708\times 6.11\sqrt{\dfrac{1}{15}+\dfrac{1}{12}}[/tex]
[-2.94 ; 5.14]
The 90% confidence interval for the difference in the mean percentage scores for the boys in calculus and the girls in calculus is [-2.94 ; 5.14].
For more information, refer to the link given below:
https://brainly.com/question/21586810
