Mathematics, 22.01.2021 19:00, shanekamayo1120

A study is run comparing HDL cholesterol levels between men who exercise regularly and those who do not. The data are shown below. Generate a 95% confidence interval for the difference in mean HDL levels between men who exercise regularly and those who do not.

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answered: 4Tris

Hello your question is incomplete attached below is the missing part of the question

Regular Exercise N Mean Std Dev

Yes 35 48.5 12.5

No 120 56.9 11.9

answer : ( -12.968, -3.832 )

Step-by-step explanation:

To generate a 95% confidence interval

first step :

Assume population variance is equal we will calculate the number of degrees of freedom

df = n1 + n2 - 2 = 35 + 120 -2 = 153

second step:

Given : critical value = 0.05 , df = 153

hence Tc = 1.976

assuming the population variance is equal calculate pooled std as ( attached below )

Sp = 12.036

third step :

calculate the standard error

Se = $Sp\sqrt{\frac{1}{n1} +\frac{1}{n2} }$

n1 = 35

n2 = 120

Sp = 12.036

hence Se = 2.312

final step

Compute the 95% confidence interval = ( -12.968, -3.832 )

attached below is the detailed solution

A study is run comparing HDL cholesterol levels between men who exercise regularly and those who do

answered: Guest

1.true 2.false? ? i think. .

step-by-step explanation:

Can somebody me ? im being timed, and i just need the main answer to this , you can explain if you

answered: Guest

$a_1 = 1$ , and

$a_{n+1} = a_n + (n + 1)$ for $n \ge 1$

step-by-step explanation:

we can analyze the pattern at which the terms increase. let us denote $a_1 = 1$ , $a_2 = 3$ , $a_3 = 6$ , so that we can call

1 the 1st term, 3 the 2nd term, 6 the 3rd term,etc.

note the following:

from the 1st term (1) to the 2nd term (3), we go up by 2,from the 2nd term (3) to the 3rd term (6), we go up by 3from the 3rd term (6) to the 4th term (10), we go up by 4, andfrom the 4th term (10) to the 5th term (15), we go up by 5.

the pattern: to get the next term value, we just add the next term number (e.g. 1st, 2nd, to the previous term value (term values are the given numbers in the question, 1, 3, 6, 10, 15,

from $a_1 = 1$ , to get to $a_2 = 3$ , we add $2(=1+1)$

from $a_2 = 3$ , to get to $a_3 = 6$ , we add $3 (=2+1)$

etc.

we need to form a recursive formula, a formula that has the next term as a function of the previous term. so a possible recursive formula can be

$a_1 = 1$ , and $a_{n+1} = a_n +(n + 1)$ for $n \ge 1$

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