Question

A cosmetic company is interested in testing the effectiveness of a new cream for treating skin blemishes. It measured the effectiveness of the new cream compared to the leading cream on the market and a placebo. Data below shows the number of blemishes removed after using the cream for 1 week. Assume two-tailed with an α = .05.

New Cream	Leading Cream	Placebo Cream
81	48	18
32	31	49
42	25	33
62	22	19
37	30	24
44	30	17
38	32	48
47	15	22
49	40	18
41	48	20

What would be my hypotheses?

What would be my critical value?

How would I compute the test statistic using ?=12?(?+1)∑??2??−3(?+1)H=12/(n(n+1))∑(T_i^2)/n_i -3(n+1) ?

Answer 1

Sol:

Null hypothesis : means of all three groups are same

Alternative hypothesis : at least one mean is different

Using Excel, (Data -> Data Analysis -> Anova: Single Factor), we get the following output -

File Home Insert Page Layout Formulas Data Review View Help ! Tell me w 127 :x r fir B C D E C F G A 1 Anova: Single Factor 2 3 SUMMARY 4 Groups 5 New Cream 6 Leading Cream 7 Placebo Cream Count Sum 10 473 10 321 10 268 Average Variance 47.3 206.6777778 32.1 113.6555556 26.8 152.1777778 8 9 10 ANOVA 11 Source of Variation 12 Between Groups 13 Within Groups 14 15 Total SS 2264.6 4252.6 df MS P-value 2 1132.3 7.189037295 0.003140596 27 157.5037037 F crit 3.354130829 6517.2 29 16 17

Now,

The value of F statistic = 7.189

and P-value = 0.00314

Since P-value < 0.05, so we reject the null hypothesis and we can conclude that new cream is more effective than the leading cream on the market and a placebo.

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