Question

In: Statistics and Probability

The experiment consisted of using three different types of chocolates: 100 g of dark chocolate (DC),...

The experiment consisted of using three different types of chocolates: 100 g of dark chocolate (DC), 100 g of dark chocolate with 200 mL of full-fat milk (DC+MK), and 200 g of milk chocolate (MC). Twelve subjects were used, 7 women and 5 men, with an average age range of 32.2±1 years, an average weight of 65.8±3.1 kg, and body-mass index of 21.9±0.4 kg m^2. On different days, a subject consumed one of the chocolate-factor levels and one hour later, the total antioxidant capacity of their blood plasma was measured. Complete the ANOVA table by using the data given below then answer the following questions. a. Does chocolate type affect the total antioxidant capacity of blood plasma? b. What chocolate type would you recommend consuming? Subjects (Observations) Factor 1 2 3 4 5 6 7 8 9 10 11 12 DC 118.8 122.6 115.6 113.6 119.5 115.9 115.8 115.1 116.9 115.4 115.6 107.9 DC+MK 105.4 101.1 102.7 97.1 101.9 98.9 100 99.8 102.6 100.9 104.5 93.5 MC 102.1 105.8 99.6 102.7 98.8 100.9 102.8 98.7 94.7 97.8 99.7 98.6

Solutions

Expert Solution

a)

From the given information,

The sample sizes under treatments are,

The total number of observations in all the samples combined is,

Note that.

The null and the alternative hypotheses to test using the ANOVA table are:

Let denote the observation under the treatment. Find the totals of the samples under the treatments.

The grand total, denoted, is the sum of all sampled items taken together.

Compute the total sum of squares, the treatment sum of squares, the error sum of squares.

Find the mean squares and the error mean square.

The value of the test statistic f is,

The degrees of freedom for the numerator and the denominator are respectively:

Under f has a distribution. From the F table, and . Since 93.58 is less than each of these values, at and at we conclude that the means are different.

The completed ANOVA table is shown below.

s f calculated value (93.58) is greater than each of these table values ( and). So, reject the null hypothesis. Hence, it can be concluded that the means are different.


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