A manufacturer wants to assess if her new weight-loss cat food is effective. She

Question

A manufacturer wants to assess if her new weight-loss cat food is effective. She has two groups of 20 adult cats that are overweight. Group 1 receives two servings of her companies regular adult formula food, while Group 2 receives two servings of her new weigh-loss formula.

The table below presents the weight loss the cats experienced over the duration of the experiment. Calculate a t value based on the two treatment groups.

Report your t-value as an absolute value (ie, positive number) to 4 decimal places.

3

Using the t-table below, and your calculated t-value for cat weight loss, you would _____ (reject or fail to reject) the null hypothesis that the weight-loss food and and regular food would have the same reduction in cat weight.

ID # Regular Weightloss 1 0 4 2 0 4 3 2 5 4 0 0 5 1 5 6 2 3 7 1 2 8 1 5 9 1 2 10 1 5 11 2 2 12 1 0 13 1 5 14 0 1 15 1 2 16 1 5 17 0 2 18 0 2 19 1 5 20 0

3

Explanation / Answer

We assume equal unknown variance for the two population. Sample size is 20 in both cases. Sample mean for regular group is=0.8 and that of weightloss group is 3.1.Variance for the two groups is 0.484 and 3.042 . So,pooled variance is=((20-1)*0.484+(20-1)*3.042)/(20+20-2)=1.763.Difference between sample average of two groups is =3.1-0.8=2.3.The test statistic is=diffrence between sample average/(pooled sd*sqrt(1/n1+1/n2))=2.3/(1.3278*sqrt(1/20+1/20))=5.478 which follows t distribution with 38 degrees of freedom.So, if we take level of signifance is= 0.05,Then upper 0.025 point oft_38 distribution(by result) is 2.0243 . |Observed t value|=5.478>2.0243. So,under 5% level of significance we reject the hypothesis that the two groups have same amount of reduction in weight.Instead we infer that,weightloss group has more weight reduction in average so that the foood has come into effect.

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