Question 12.2: Tall Buildings A researcher wishes to see if there is a diff......
Tall Buildings
A researcher wishes to see if there is a difference in the number of stories in the tall buildings of Chicago, Houston, and New York City. The researcher randomly selects five buildings in each city and records the number of stories in each building. The data are shown. At 𝛼 = 0.05, can it be concluded that there is a significant difference in the mean number of stories in the tall buildings in each city?
\begin{array}{|c|c|c|} \hline \text{Chicago }& \text{Houston }& \text{New York City} \\ \hline 98 & 53 & 85 \\ 54 & 52 & 67 \\ 60 & 45 & 75 \\ 57 & 41 & 52 \\ 83 & 36 & 94 \\ 49 & 34 & 42 \\ \hline \end{array}
Source: The World Almanac and Book of Facts
Step 1 State the hypotheses and identify the claim.
H_{0}: \mu_{1}=\mu_{2}=\mu_{3}
H_{1} : At least one mean is different from the others (claim).
Step 2 Find the critical value. Since k=3, N=18, and \alpha=0.05,
d.f.N. =k-1=3-1=2
d.f.D. =N-k=18-3=15
The critical value is 3.68.
Step 3 Compute the test value.
a. Find the mean and variance of each sample. The mean and variance for each sample are
Chicago \quad \bar{X}_{1}=66.8 \quad s_{1}^{2}=371.8
Houston \bar{X}_{2}=43.5 \quad s_{2}^{2}=63.5
New York \bar{X}_{3}=69.2 \quad s_{3}^{2}=387.7
b. Find the grand mean.
\bar{X}_{\mathrm{GM}}=\frac{\Sigma X}{N}=\frac{98+54+60+\cdots+42}{18}=\frac{1077}{18}=59.8
c. Find the between-group variance.
\begin{aligned}s_{B}^{2} & =\frac{\sum n_{i}\left(\bar{X}_{i}-\bar{X}_{\mathrm{GM}}\right)^{2}}{k-1} \\& =\frac{6(66.8-59.8)^{2}+6(43.5-59.8)^{2}+6(69.2-59.8)^{2}}{3-1} \\& =\frac{2418.3}{2}=1209.15\end{aligned}
d. Find the within-group variance.
\begin{aligned}s_{W}^{2} & =\frac{\Sigma\left(n_{i}-1\right) s_{i}^{2}}{\Sigma\left(n_{i}-1\right)} \\& =\frac{(6-1)(371.8)+(6-1)(63.5)+(6-1)(387.7)}{(6-1)+(6-1)+(6-1)}=\frac{4115}{15} \\& =274.33\end{aligned}
e. Find the F test value.
F=\frac{s_{B}^{2}}{s_{W}^{2}}=\frac{1209.15}{274.33}=4.41
Step 4 Make the decision. Since 4.41>3.68, the decision is to reject the null hypothesis. See Figure 12-2.
Step 5 Summarize the results. There is enough evidence to support the claim that at least one mean is different from the others. The ANOVA summary table for this example is shown in Table 12-3.
In this case, 4.41 falls between 4.77 and 3.68, which corresponds to 0.025 at the 0.05 level. Hence, 0.025<P-value <0.05. Since the P-value is less than 0.05, the decision is to reject the null hypothesis. The P-value obtained from the calculator is 0.031.
| TABLE 12–3 Analysis of Variance Summary Table for Example 12–2 | ||||
| Source | Sum of squares | d.f. | Mean square | F |
| Between | 2418.3 | 2 | 1209.15 | 4.41 |
| Within | 4115 | 15 | ||
| Total | 6533.3 | 17 | ||
