Question

Some individuals believe that there is a link between mobile phone use and brain tumors. The basic idea is that mobile phones emit microwaves, and so holding one next to your brain for large parts of the day is a bit like sticking your brain in a microwave oven and selecting the ‘cook until well done’ button. (Please note: this claim has been refuted. But if it were supported, the study described below would not be a realistic or ethical way to test this claim.) Hypothetically, if researchers were to test this experimentally, they could get six groups of people and strap a mobile phone to their heads (that they can’t remove). Then, by remote control, they turn the phones on for a certain amount of time each day. After 6 months, they could measure the size of any tumor (in mm2 ) close to the site of the phone. The six groups experienced 0, 1, 2, 3, 4, or 5 hours per day of phone microwaves for 6 months. 1.) State the null and alternative hypothesis for this study. Do so as both a sentence and using the appropriate mathematical representation (uncommon symbols to copy and edit as needed: σµ 2 ). 2.) Using the procedure for conducting a one-way ANOVA, run the analysis placing the independent and dependent variables in their appropriate dialog boxes. In addition to the one-way ANOVA, conduct a Tukey’s HSD post-hoc test. That data will be used to answer question 3. Copy and paste the output of your ANOVA in your MSWord document (under #1). 3.) Although an ANOVA can tell us whether at least one group differs from all other groups, it does not tell us which groups are different. To determine this outcome we use a post-hoc test. Based on the results of your Tukey’s HSD post-hoc comparison, indicate which groups, if any, were significantly different from one another in this analysis. 4.) Based on a significance level of α = 0.05, report your results in APA format. 5.) In a couple of sentences, explain what the results of this study mean. Think of it as if you were trying to explain the meaning of the statistical analysis to your grandparents. How would you describe the results to them?

.00   .04
.00   .02
.00   .03
.00   .03
.00   .06
.00   .06
.00   .03
.00   .05
.00   .02
.00   .03
.00   .03
.00   .04
.00   .05
.00   .04
.00   .03
.00   .06
.00   .05
.00   .04
.00   .03
.00   .03
1.00   .79
1.00   .76
1.00   .24
1.00   .96
1.00   .64
1.00   .35
1.00   .49
1.00   .80
1.00   .78
1.00   .05
1.00   .95
1.00   .41
1.00   .64
1.00   .50
1.00   .17
1.00   .74
1.00   .40
1.00   .29
1.00   .02
1.00   .71
2.00   1.31
2.00   1.10
2.00   1.09
2.00   .50
2.00   1.28
2.00   .54
2.00   .66
2.00   1.71
2.00   1.87
2.00   1.24
2.00   1.71
2.00   2.36
2.00   1.56
2.00   1.89
2.00   .85
2.00   1.00
2.00   1.00
2.00   1.41
2.00   1.65
2.00   .89
3.00   4.33
3.00   2.49
3.00   2.06
3.00   3.34
3.00   3.20
3.00   2.26
3.00   3.82
3.00   2.88
3.00   2.60
3.00   4.11
3.00   3.53
3.00   3.04
3.00   3.65
3.00   4.16
3.00   2.84
3.00   1.79
3.00   2.58
3.00   3.61
3.00   2.66
3.00   1.86
4.00   4.67
4.00   5.18
4.00   4.08
4.00   4.63
4.00   5.34
4.00   4.86
4.00   5.22
4.00   5.42
4.00   3.06
4.00   4.75
4.00   5.20
4.00   5.11
4.00   5.67
4.00   3.85
4.00   4.90
4.00   5.23
4.00   5.32
4.00   6.07
4.00   4.15
4.00   5.44
5.00   5.19
5.00   5.05
5.00   6.16
5.00   4.92
5.00   4.67
5.00   3.90
5.00   5.27
5.00   2.72
5.00   5.33
5.00   5.38
5.00   5.45
5.00   4.45
5.00   4.85
5.00   4.15
5.00   4.07
5.00   3.69
5.00   4.45
5.00   4.64
5.00   5.13
5.00   5.57

Answer 1

here we use F-test using one-way anova with

(1) null hypothesis H0:0hour=1hour=2hour=3hour=4hour=5hour

and alternate hypothesis H0: at least one hour is different from others.

(2) since p-value= 0.000 l between the group is less than typical level of significance alpha=0.05, so we reject null hypothesis and now we proceed for post-hoc test for multiple comparison to find which pair is different to each other.

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
0 hours	20	0.77	0.0385	0.000161
1 hours	20	10.69	0.5345	0.081342
2 hours	20	25.62	1.281	0.241778
3 hours	20	60.81	3.0405	0.587679
4 hours	20	98.15	4.9075	0.483651
5 hours	20	95.04	4.752	0.610996
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	450.6936	5	90.13873	269.6603	2.04E-61	2.293911
Within Groups	38.10652	114	0.334268
Total	488.8001	119

(3) since the n1=10 and n2=10 for all the pair , it is same. so for each pair wise comparison  HSD is same

HSD=0.7496 ( using two decimal place approximation

Tukey HSD=	q*sqrt((MSE/2)*(1/n1+1/n2)=	0.7496
q=	4.1
MSE=	0.3343
n1=	10
n2=	10

(4) please find the details of pair-wise comparison

comparison		difference	HSD	significant
0hour	1hour	0.4960	0.7496	not
0hour	2hour	1.2425	0.7496	yes
0hour	3hour	3.0020	0.7496	yes
0hour	4hour	4.8690	0.7496	yes
0hour	5hour	4.7135	0.7496	yes
1Hour	2hour	0.7465	0.7496	not
1Hour	3hour	2.5060	0.7496	yes
1Hour	4hour	4.3730	0.7496	yes
1Hour	5hour	4.2175	0.7496	yes
2hour	3hour	1.7595	0.7496	yes
2hour	4hour	3.6265	0.7496	yes
2hour	5hour	3.4710	0.7496	yes
3hour	4hour	1.8670	0.7496	yes
3hour	5hour	1.7115	0.7496	yes
4hour	5hour	0.1555	0.7496	not