Question

2. A statistic professor wants to investigate that there is a difference between a man's height and a woman's height at SMC. She gathered 23 female statistics students in her class and found their mean height is 64.217 inches and their standard deviation is 2.2755 inches. On the other hand, there were 12 male statistics students who were asked their height; their mean height is 67.833 inches and the standard deviation is 3.738 inches.  

Construct a 95% confidence interval estimate for the difference between the corresponding population mean height of gender at SMC. (10 points)

• Find the point estimate (the difference between sample means)
• Determine critical value
• Find the margin of error
• Construct a confidence interval
• Does it appear that there is a difference in their height between men and women?

Answer 1

Given :-

There were 12 male statistics students who were asked their height; their mean height is 67.833 inches and the standard deviation is 3.738 inches.  

So   n1 = 12 ;    1 = 67.833    ;    s1= 3.738

23 female statistics students in her class and found their mean height is 64.217 inches and their standard deviation is 2.2755 inches

Thus   n2 = 23 ;    2 = 64.217    ;    s2 = 2.2755

Question . Construct a 95% confidence interval estimate for the difference between the corresponding population mean height of gender at SMC.

i) Find the point estimate (the difference between sample means)

Point estimate is given by 1 - 2 = 67.833 - 64.217 = 3.616

Thus point estimate = 3.616

[Note that:- this point estimate is difference between sample means of male height and female height respectively].

ii) Determine critical value

To find t-critical value .

Here is t-distributed with n1+n2-2 = 12+23-2 = 33 degree of freedom

We have = 0.05 [ for 95% confidence ]

can be obtained from statistical book or from any software like R/Excel .

From R

> qt(1-0.05/2,df=33)
[1] 2.034515

Thus = 2.034515

So critical value is 2.034515

iii) Find the margin of error

Margin of error is given by

ME = * Se

Now Se =

where sd² =

Calculation :-

sd² =

        = ( (12-1 )*3.738² + (23-1)* 2.2755² ) / (12+23-2)

sd² = 8.109481

Se =

      =

Se = 1.014089

and = 2.034515

Hence ME = * Se

              ME = 2.034515 * 1.014089

Margin of error    ME = 2.063179

iv)Construct a confidence interval

95% confidence inteval is given by

CI = { 1 - 2 - ME , 1 - 2 + ME }

     = { 3.616 - 2.063179 , 3.616 + 2.063179 }

CI = { 1.552821 , 5.679179 }

Thus 95% confidence interval is { 1.552821 , 5.679179 }

v) Does it appear that there is a difference in their height between men and women?

The above calculated confidence interval is { 1.552521 , 5.67179 }

Note that zero " 0 " is not included in above interval which implies that there is a difference in their heights between men and women .

Since both lower bound and upper bound of interval are greater that 0 , we can say that height of male amy be significantly greater than that of females.

Thus , Yes , There is a difference in their height between men and women .