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..

1. 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

2. 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.