Question

13. Jordan's cat --Fernis a finicky eater Jordan is trying to determine which of two brands of canned catfood Fern prefers Tab-a-Cator Chow Lion. For two months, she flipsa coin each day to decide which of the two foods to feed Fern, and weighs how much Ferneatsis grams. Here is the data 2 gample T- Test Tab-a-Cat Chow Lion 31 1852 13.45 30 82.14.62 (a) Construct an appropriate significance testata 05 alpha level to determine if there is a significant difference in Fera's weight based on the brand of cat food er is eating Hoil, = M2 2 sample T-Test at d= .05 Mail EM TE $5.2 - 82.1) - 3.1 1.047 = 3..-12.960 P=.005 Conditions: Random: Can flup - Normal N 30 Indep: yes Reject the null hypothesis blc pca. There is a Significance difference in how much fern likes tab-a-cut a chowian. pob 007 (b) Construct and interpret when she is offered Ow en infidence interval for the difference in mean amount of food Ferneats when she is offered Tabs Cat and 99% confident that the actual difference in Fera's weight is between .214 and 5.966 pounds after 2 SAHS. 3.1 + 2.756 3.11 2.756 (1.047) 3.1 2.886 |6.214,5.966) #TFI...

Answer 1

Since standard deviations are different, we use 2- sample t-test with unequal variance

The following null and alternative hypotheses need to be tested: Ho: Mi = u2 Ha: Mi 7 u2 This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the degrees of freedom are df = 53.644. In fact, the degrees of freedom are computed as follows, assuming that the population variances are unequal: Hence, it is found that the critical value for this two-tailed test is te = 2.005, for a = 0.05 and df = 53.644. The rejection region for this two-tailed test is R = {t:t > 2.005). (3) Test Statistics Since it is assumed that the population variances are unequal, the t-statistic is computed as follows: t= X1 – X2 si si V ning 85.2 – 82.1 = 2.962

(4) Decision about the null hypothesis Since it is observed that t = 2.962 > t = 2.005, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p = 0.0045, and since p = 0.0045 < 0.05, it is concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population mean jy is different than 42, at the 0.05 significance level.

b)

99% confidence interval

df = 53.644

t for 99% confidence level = 2.671

Confidence Interval The 99% confidence interval is 0.305<Mi - M2 < 5.895.