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Answers without the blur. Sign up and see all textbooks for free! Q. 101

Expert-verified Found in: Page 604 ### Introductory Statistics

Book edition OER 2018
Author(s) Barbara Illowsky, Susan Dean
Pages 902 pages
ISBN 9781938168208 # 101. Elizabeth Mjelde, an art history professor, was interested in whether the value from the Golden Ratio formula, $\left(\frac{\text{larger + smaller dimension}}{\text{larger dimension}}\right)$ 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?

Subscripts : $1=1900$ to $1919,2=1920$ to $1942$.
(a) The null hypothesis: ${H}_{0}:{\mu }_{1}={\mu }_{2}$(b) The alternate hypothesis: ${H}_{0}:{\mu }_{1}\ne {\mu }_{2}$(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) The test statistics $-0.27$.
(f) The $p$-value is $0789$.(g)(i) $\alpha =0.05$(ii) Decision: Do not reject the null hypothesis.(iii) Reason for Decision: $p$ - value $>\alpha$.(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.

See the step by step solution

## Step 1: Given information

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 $0.1064$.

## Step 2: Explanation

Subscripts :$1=1900$ to $1919,2=1920$ to $1942$.
(a) The null hypothesis is determined as:${H}_{0}:{\mu }_{1}={\mu }_{2}$(b) The alternate hypothesis is determined as:${H}_{0}:{\mu }_{1}\ne {\mu }_{2}$(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 $-0.27$.

## Step 3: Explanation

(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) $\alpha =0.05$
(ii) Decision: Do not reject the null hypothesis.(iii) Reason for Decision: $p$- value $>\alpha$.(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. ### Want to see more solutions like these? 