Elizabeth Mjelde an art history professor was interested in

Elizabeth Mjelde, an art history professor, was interested in whether the value from the Golden Ratio formula, ( larger + smaller dimension) / larger dimension ) was the same in the Whitney Exhibit for works from 1900 to 1919 as for works from 1920 to 1942. Thirty-seven early works were sampled, averaging 1.74 with a standard deviation of 0.11. Sixty-five of the later works were sampled, averaging 1.746 with a standard deviation of 0.1064. Do you think that there is a significant difference in the Golden Ratio calculation?

Solution

We have given that ,Elizabeth Mjelde, an art history professor, was interested in whether the value from the Golden Ratio formula, ( larger + smaller dimension) / larger dimension ) was the same in the Whitney Exhibit for works from 1900 to 1919 as for works from 1920 to 1942.

Consider alpha = level of significance = 0.05

Here hypothesis for the test is,

H0 : µ1 = µ2 Vs H1 : µ1   µ2

The test is two sided.

Here first we have to check whether the two variances are equal or not.

Hypothesis for the test is,

H0 : The two variances are equal.

H1 : The variances are not equal.

The test statistic is,

F = Larger variance / Smaller variance

We have given that,

n1 = number of early works = 37

n2 = number of later works = 65

X1bar = 1.74

X2bar = 1.746

s1 = 0.11

s2 = 0.1064

F = 0.11^2 / 0.1064^2 = 1.0688

critical value for F distribution we can calculate by using EXCEL.

syntax :

=FINV(probability , d.f.1, d.f.2)

probability = alpha / 2 = 0.05 / 2 = 0.025

d.f.1 = n1 - 1 = 37 - 1 = 36

d.f.2 = n2 - 1 = 65 - 1 = 64

Critical value = 1.7523

F < critical value

Fail to reject H0 at 5% level of significance.

Conclusion : Variances are equal.

Now we test two means using pooled variance.

Pooled variance (S) = sqrt [ (n1-1) * s1^2 + (n2 - 1) * s2^2 / (n1 + n2 -2) ]

S = sqrt [ (37-1) * 0.11^2 + (65-1) * 0.1064^2 / 100 ]

= sqrt(1.1601 / 100)

S = 0.10771

The test statistic is,

t = (X1bar - X2bar) / [ S * sqrt(1/n1 + 1/n2) ]

= ( 1.74 - 1.746) / [ 0.10771 * sqrt ( 1 / 37 + 1 / 65 ) ]

t = - 0.006 / 0.0222 = - 0.27049

t = - 0.27049

We can find P-value for taking decision.

EXCEL syntax :

=tdist ( x,d.f.,tails)

x is the test statistic value

d.f. = n1 + n2 - 2 = 37 + 65 - 2 = 100

tails = 2

P-value = 0.787341

P-value > alpha

Fail to reject H0 at 5% level of significance.

Conclusion : The two means are equal.