Question

Aresearcher wants to determine the impact that smoking has on resting heart rate. She randomly selects 7 people from 3 groups and obtaina the heart rate data (beats per minute) in the table. Complete parts (a) through (c). Nonsmokers (a) Test the null hypothesis that the mean resting heart rate for each category e at the a - 0.05 level of significance Nate that the ay ANOVA are saisfied ldertty the nulland atemate hvectheses OA O B O C .: At leact one of the means is differant dentfy the Pyalue. P-value Round to three decimal places as neadad Choose the correct conclusion below OA Since P-value> do not reject the r hypothesis reject the null hypothesis . since Pvalue s hesis. OD Since P value» ,reject the null hypothesis f the nul hypothesis is rejected in part (a), use Tukey's test to determine which pairwise means differ using a familywise eror rate of a-0.05 Find the Tukey 95% simultaneous oconidence intervals for all painwise comparisons Select the conect choice below and if necessary, fill in any answer boxes to complete your choice b) (Round to two decimal places as needed) OA. The interval for the paiise compariaion - The interval for the painaise comparsion i or rep ersion - OB. ul hypothesis was not rejected. State the conolusion the hypothesis He: the hypothesis H te hypethesis He: s s " and ShoDER the oet seu re treatment levels t0 suppot the analytic resuts obtained in pans (a) and () O B OA LS IT 18 NS 4 00 00r Pbo

Answer 1

Treatments
	1	2	3	Total
N	7	7	7		21
∑X	387	465	521		1373
Mean	55.2857	66.4286	74.4286		65.381
∑X2	21693	31081	39019		91793
Std.Dev.	7.0407	5.6526	6.3471		10.0622

a)

The test hypotheses are,

H0: $\mu12 = \mu13 = \mu23$

Ha: At least one of the mean is different.

Option -(C)

ANOVA Table:

Source	SS	df	MS
Between-treatments	1294.0952	2	647.0476	F = 15.93589
Within-treatments	730.8571	18	40.6032
Total	2024.9524	20

The f-ratio value is 15.93589. The p-value is .000104.

The result is significant at p < .05.

Since the p-value < $\alpha$. Hence reject the null hypothesis H0.

Option -(B)

b)

Tukey's test:

Q statistic at α = 0.05, k = 3, N-k = 18 , Q = 3.61

Critical range, CV = Q* √(MS(within)/n) =  9.2844

Comparison	Absolute Diff. = |xi - xj|	Critical Range, CV = Q* √(MSW/n)	Confidence interval
			|xi - xj| - CV	|xi - xj| + CV
µ2-µ1	11.1429	9.2844	1.8585	20.4273
µ3-µ1	19.1429	9.2844	9.8585	28.4273
µ3-µ2	8	9.2844	-1.2844	17.2844

The hypothesis H0: $\mu12 \neq \mu13$

The hypothesis H0: $\mu13 \neq \mu23$

the hypothesis H0: $\mu12 \neq \mu23$