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3.10 Corny Flavors 5 pts. Suppose you work for the Frito-Day serpecation in their new foods development division. They have d
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Answer #1

An One factor ANOVA is the most suitable analysis for the above problem. ie to compare three or more groups.

The null hypothesis is that verifresh has not affected the taste appeal in tortilla chips vs the alternative that verifresh affect the taste appeal in tortilla chips.

The degrees of freedom are (no: of groups-1)=3-1=2; and ( total observation -no:of groups)=90-3=87

The critical value is for (2,87) degrees of freedom is 3.11 from the F table for 0.05 significance level or alpha. The statistic should follow F distribution which takes values from 0 to infinity. so we call the statistic, F statistic, which is the ratio of two variances , that cannot be negative. so the F statistic will always take value >0. So I take the magnitude of the statistic value ie 2.5. Since 2.5 < critical value(3.11) we cannot reject the null hypothesis at 5% alpha that the verifresh has not affected the taste appeal in tortilla chips.

A two sample t test would be appropriate in this case here we test whether pearl stimulant has an effect on the size of pearl produced which is considered to be our alternative hypothesis. where the null hypothesis is pearl stimulant has no effect on the size of pearl produced. The test is one tailed as we are interested in knowing our treatment is effective.

The value of t statistic is 1 and the critical value for 98 degrees of freedom and 5% alpha is 1.661

SurfStat t-distribution calculator -t 0t t 0 vil probability d.f. t value 0.05 1.661 98

Since t statistic <1.661 we cannot reject the null hypothesis. Hence we conclude that the treatment is not effective.

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3.10 Corny Flavors 5 pts. Suppose you work for the Frito-Day serpecation in their new foods development division. They have developed a new additive they've called Veriixesh. They want to add...
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