Question

1. Betty wants to know if eatine chocolate makes people laugh more. To that end, she designs astu where people are divided into three groups - one group eats no chocolate, one group eats a chocolate and one group eats a lot of chocolate, Betty recruits 15 people for her study and random assigns each person to one of the three ground The 15 people spend a week in their respective groups and then all people are shown the same comedic video. Betty records how much time each person spenas laughing. The data follows. (HINT: we used this question in the bonus review for exam 1 so you probably already have some things computed.) No chocolate Bit of chocolate Lots of chocolate 20 seconds 30 seconds 45 seconds 30 seconds 28 seconds 48 seconds 25 seconds 15 seconds 47 seconds 10 seconds 29 seconds 60 seconds 14 seconds 33 seconds 80 seconds Does eating chocolate make people laugh significantly more? (i.e., what can Betty conclude?)

Answer 1

I used MINITAB to solve this question.

Step.1 Enter data in MINITAB. Go to 'Stat' menu ----> ANOVA ----> One Way ANOVA . New window will pop up on screen.

Minitab - Untitled Eile Edit Data 10% Calc Session Stat Graph Editor Tools Basic Statistics Regression ANOVA DOE Control Charts Quality Tools Reliability/Survival Multivariate Time Series Tables Nonparametrics Equivalence Tests Power and Sample Size Window Help Assistant OESG GOODSOOG TvA OTOO LUM One-Way... # A One-Way 9 Determine whether the means of two G or more groups differ. Fully Nested ANOVA... General MANOVA... 02 Test for Equal Variances... Interval Plot... Main Effects Plot... Interaction Plot... A C4 C5 C6 C7C809010 Worksheet 1 *** C1 No 20 - 30 25 10 14 C2 Bit 30 28 15 29 33 C3 Lots 45 48 47 60 80

Step. 2

Select appropriate option in the first row. Then select response. then go to comparison option select 'Dunnet test' and control group level equal to 'Lots' and press OK.

Minitab - Untitled Eile Edit Data Calc Stat Graph Editor Tools Window Help Assistant #HO%03 SO GOOD SOD GIF | -- DIA 13 32 NA+Y I XIQTOO. IM Session One-Way Analysis of Variance: Comparisons X One-Way Analysis of Variance Error rate for comparisons: 5 Response data are in a separate column for each factor level ci No C2 Bit C3 Lots Comparison procedures assuming equal variances Tukey Responses: No-Lots Fisher Dunnett Control group level: Lots Hsu MCB Best: Largest mean is best Results Options... Comparisons... Graphs... Worksheet 1 *** 01 No - 20 30 02 Bit 30 28 03 Lots 45 48 Interval plot for differences of means Grouping information Tests Results... Select Storage... Help OK Cancel 29 60 Help OK Cancel 33

Output:

One-way ANOVA: No, Bit, Lots

Method

Null hypothesis All means are equal
Alternative hypothesis At least one mean is different
Significance level α = 0.05

Equal variances were assumed for the analysis.

Factor Information

Factor Levels Values
Factor 3 No, Bit, Lots

Analysis of Variance

Source DF Adj SS Adj MS F-Value P-Value
Factor 2 3672 1836.1 16.78 0.000
Error 12 1313 109.4
Total 14 4985

Model Summary

S R-sq R-sq(adj) R-sq(pred)
10.4594 73.66% 69.28% 58.85%

Means

Factor N Mean StDev 95% CI
No 5 19.80 8.07 ( 9.61, 29.99)
Bit 5 27.00 6.96 (16.81, 37.19)
Lots 5 56.00 14.65 (45.81, 66.19)

Pooled StDev = 10.4594

Dunnett Multiple Comparisons with a Control

Grouping Information Using the Dunnett Method and 95% Confidence

Factor N Mean Grouping
Lots (control) 5 56.00 A
Bit 5 27.00
No 5 19.80

Means not labeled with the letter A are significantly different from the control level mean.

Conclusion:

P-value for F test is 0.000. Hence we reject null hypothesis and conclude that at least one mean is significantly different than other.

We use Dunnet test for comparing mean of Lots of chocolate group to other group. We see that these are not equal.

and average of 'Lots of chocolate' group is greater than others.

Hence we conclude that eating chocolate make people laugh significantly more.