91Ó°ÊÓ

Let, the thirty-seven early works were sampled, averaging $1.74$with a standard deviation of $0.11$. And the sixty-five of the later works were sampled, averaging $1.746$with a standard deviation of $0.1064$.

Subscripts :$1=1900$to $1919,2=1920$to $1942$.
(a) The null hypothesis is determined as:
$H_0:{\mathrm{μ}}_1={\mathrm{μ}}_2$
(b) The alternate hypothesis is determined as:
$H_0:{\mathrm{μ}}_1≠{\mathrm{μ}}_2$
(c) The random variable is the difference between the means of the golden ratio formula for works from 1900 to 1919 and works from 1920 to 1942 in the Whitney Exhibit.
(d) Student's t distribution.
(e) Using Minitab's two sample t test option is determined as:

Hence, the test statistics $-0.27$.

(f) The $p$-value from the output is determined as $0789$.
(g) Plot a graph of the region(s) that correspond to the $p$-value.

(i) $\mathrm{Î\pm }=0.05$
(ii) Decision: Do not reject the null hypothesis.
(iii) Reason for Decision: $p$- value $>\mathrm{Î\pm }$.
(iv) As a result: There is insufficient information to infer that the means of the golden ratio formula for Whitney Exhibit pieces from 1900 to 1919 and works from 1920 to 1942 are different at the $5\%$significance level.