Question

A random sample of 25 wolf litters in Ontario, Canada, gave an average of X1 = 5.6 wolf pups per litter, with estimated sample standard deviation S1 = 0.9. Another random sample of 14 wolf litters in Finland gave an average of x2 = 2.2 wolf pups per litter, with sample standard deviation s2 = 0.8. (a) Categorize the problem below according to parameter being estimated, proportion p, mean u, difference of means uu - U2, or difference of proportions P1 - P2. Then solve the problem. OP O M1 - H2 O P1 - P2 ou (b) Find an 95% confidence interval for M1 - U2, the difference in population mean litter size between Ontario and Finland. (Use 1 decimal place.) lower limit upper limit (c) Examine the confidence interval and explain what it means in the context of this problem. Does the interval consist of numbers that are all positive? all negative? of different signs? At the 95% level of confidence, does it appear that the average litter size of wolf pups in Ontario is greater than the average litter size in Finland? O Because the interval contains only positive numbers, we can say that the average litter size of wolf pups is greater in Ontario. o Because the interval contains both positive and negative numbers, we can not say that the average litter size of wolf pups is greater in Ontario. o We can not make any conclusions using this confidence interval. O Because the interval contains only negative numbers, we can say that the average litter size of wolf pups is greater in Finland.

Answer 1

solution: a) we need to test the difference between means hence second option is connect ou la b) Given x = 5.6 5,=0.9, 0,=25 Xq=2.2, Sq=0.8, N2 = 14 :1-95% = 1-0.99=0.09 dfa nitha -2=25+14-2-37 o cocitical = tollesdf = to.os, 37 = 2.00 Pooled estimate of variance is goe (1,-1) 5,²+ (na 18² (9,-1)(not 25-1) 0.9°4(14-130.92 125-1)+(14-13 +0.750 27027027 Sp 0.7503703 Standard er der Säit 0275276534 0.7502703 -0.28913932232 ~ 0- 28913932 margin of Cavour -(£)= tolo. of * tot 2.026 X0-28913932 -0.585796262 20.585796 Point of estimated difference = K X = 5.6-2.2= 3.4 comits of 95% CI are given by lower limit -13,-2) - € = 3.4-0.585796--2.814 -2.8 Upper Cmite ngła) +E=3.4-0-58 5796 = 3.985 24 25

c) Because the internal contains onby posizue numbers, we can say that the average littersize of wolf Purs is greater in ontario option first is coororect.
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