Question

An investigator in the Statistics Department of a large university is interested in the effect of exercise in maintaining mental ability. She decides to study the faculty members aged 40 to 50 at his university, looking separately at two groups: The ones that exercise regularly, and the ones that don’t. There turn out to be several hundred people in each group, so she takes simple random sample of 25 persons from each group, for detailed study. One of the things she does is to administer an IQ test to the sample people, with the following results:

Regular Exercise:

Sample size: 25

Average score: 130

Standard deviation: 15

No Regular Exercise:

Sample size: 25

Average score: 120

Standard deviation: 15

The investigator concludes that exercise does indeed help to maintain mental ability among the faculty members aged 40 to 50 at his university. Is this conclusion justified? Explain whether you agree with her and show your reasoning mathematically

Answer 1

Result:

n investigator in the Statistics Department of a large university is interested in the effect of exercise in maintaining mental ability. She decides to study the faculty members aged 40 to 50 at his university, looking separately at two groups: The ones that exercise regularly, and the ones that don’t. There turn out to be several hundred people in each group, so she takes simple random sample of 25 persons from each group, for detailed study. One of the things she does is to administer an IQ test to the sample people, with the following results:

Regular Exercise:

Sample size: 25

Average score: 130

Standard deviation: 15

No Regular Exercise:

Sample size: 25

Average score: 120

Standard deviation: 15

The investigator concludes that exercise does indeed help to maintain mental ability among the faculty members aged 40 to 50 at his university. Is this conclusion justified? Explain whether you agree with her and show your reasoning mathematically

Two sample t test used.

Ho: µ₁ = µ₂   H₁: µ₁ > µ₂

Upper tail test used.

= 2.3570

DF = n₁+n₂-2 =48

Table value of t with 48 DF at 0.05 level = 1.6772

Rejection Region: Reject Ho if t > 1.6772

Calculated t = 2.3570, falls in the rejection region

The null hypothesis is rejected.

There is enough evidence to conclude that exercise does indeed help to maintain mental ability among the faculty members aged 40 to 50 at his university.

Pooled-Variance t Test for the Difference Between Two Means
(assumes equal population variances)
Data
Hypothesized Difference	0
Level of Significance	0.05
Population 1 Sample
Sample Size	25
Sample Mean	130
Sample Standard Deviation	15
Population 2 Sample
Sample Size	25
Sample Mean	120
Sample Standard Deviation	15
Intermediate Calculations
Population 1 Sample Degrees of Freedom	24
Population 2 Sample Degrees of Freedom	24
Total Degrees of Freedom	48
Pooled Variance	225.0000
Standard Error	4.2426
Difference in Sample Means	10.0000
t Test Statistic	2.3570
Upper-Tail Test
Upper Critical Value	1.6772
p-Value	0.0113
Reject the null hypothesis