Mathematics: Astudent researcher compares the heights of men and women from the student body of a certain college in order to estimate the difference in their mean heights. a random sample of 15 men had a mean height of 69.4 inches with a standard deviation of 3.09 inches. a random sample of 7 women had a mean height of 64.9 inches with a standard deviation of 2.58 inches. determine the 99% confidence interval for the true mean difference between the mean height of the men and the mean height of the women. assume

Astudent researcher compares the heights of men

Ответ:

alessandrotabares

04.01.2022

Mathematics

(0.6613, 8.3387)

Step-by-step explanation:

Let the heights of men be the first population and the heights of women be the second population. Then

We have $n_{1} = 15$ , $\bar{x}_{1} = 69.4$ , $s_{1} = 3.09$ and

$n_{2} = 7$ , $\bar{x}_{2} = 64.9$ , $s_{2} = 2.58$ . The pooled

estimate is given by

$s_{p}^{2} = \frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(15-1)(3.09)^{2}+(7-1)(2.58)^{2}}{15+7-2} = 8.68$

The 99% confidence interval for the true mean difference between the mean height of the men and the mean height of the women is given by (if the samples are independent)

$(\bar{x}_{1}-\bar{x}_{2})\pm t_{0.01/2}s_{p}\sqrt{\frac{1}{15}+\frac{1}{7}}$ , i.e.,

$(69.4-64.9)\pm t_{0.005}2.946\sqrt{\frac{1}{15}+\frac{1}{7}}$

where $t_{0.005}$ is the 0.5th quantile of the t distribution with (15+7-2) = 20 degrees of freedom. So

$4.5\pm(-2.845)(2.946)(0.458)$ , i.e.,

(0.6613, 8.3387)

0,0(0 оценок)

Ответ:

MengXiong1

23.02.2023

Mathematics

AROC =6

Step-by-step explanation:

Let f(x)=6x-9

When x=2

f(2) =6(2)-9 = 3

When x=0

f(0)=6(0)-9 =-9

So, average rate of change AROC= $\frac{3-(-9)}{2-0} =\frac{12}{2}$ =6

So, AROC=6

0,0(0 оценок)

Ответ:

Ответ:

Ask your question to the AI advisor

More tips

Answers on questions: Mathematics

Ask an AI advisor a question