Question

A researcher wants to determine the impact that smoking has on resting heart rate. She randomly selects

77

people from

33

groups and obtains the heart rate data​ (beats per​ minute) in the table. Complete parts ​(a) through

​(c).

Nonsmokers	5050	5050	4848	6262	6767	5252	4747	Open in StatCrunch + Copy to Clipboard + Open in Excel +
Light Smokers	7272	6060	6565	6969	6363	7373	5959
Heavy Smokers	7171	8383	6363	7878	7575	7676	7676

​(a)	Test the null hypothesis that the mean resting heart rate for each category is the same at the alpha equals 0.05α=0.05 level of significance. Note that the requirements for a​ one-way ANOVA are satisfied.

Identify the null and alternative hypotheses.

A.

Upper H 0H0​:

mu Subscript ns Baseline equals mu Subscript l s Baseline equals mu Subscript hsμns=μls=μhs

Upper H 1H1​:

mu Subscript ns Baseline not equals mu Subscript l s Baseline not equals mu Subscript hsμns≠μls≠μhs

Answer 1

Solution:

a. The null and alternative hypotheses are:

We can use the excel ANOVA: Single Factor data analysis tool to find the answer to the given questions.
The excel steps are:

Enter the data in excel.

Click on Data > Data Analysis > ANOVA: Single Factor > OK

Input Range: Select the data range for all the data including labels

Mark Labels in the first row

Alpha = 0.05

Choose the output range and click OK.

The output is given below:

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
Nonsmokers	7	376	53.71428571	58.9047619
Light smokers	7	461	65.85714286	31.47619048
Heavy Smokers	7	522	74.57142857	38.95238095
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	1536.285714	2	768.1428571	17.81774669	0.0001	3.554557146
Within Groups	776	18	43.11111111
Total	2312.285714	20

Since the p-value is less than the significance level, we, therefore, reject the null hypothesis and conclude that the mean resting heart rate for each category is not the same at the 0.05 significance level.