Question

Q. 1 A researcher wishes to see if the average length of the major rivers in the United States is the same as the average length of the major rivers in Europe. The data (in miles) of a random sample of rivers are shown. At α = 0.01, is there enough evidence to reject the claim? Assume σ1 = 450 and σ2 = 474. (Marks = 6)

USA			Europe			USA			Europe
729	560	434	481	724	820	1243	605	360	447	567	252
329	332	360	532	357	505	525	926	722	824	932	600
450	2315	865	1776	1122	496	850	310	430	634	1124	1575
330	410	1036	1224	634	230	532	375	1979	565	405	2290
329	800	447	1420	326	626	710	545	259	675	454
600	1310	652	877	580	210	300	479	425

Answer 1

The hypothesis being tested is:

H₀: µ₁ = µ₂

H₁: µ₁ ≠ µ₂

USA	Europe
682.6666667	810.2857143	mean
450	474	std. dev.
18	14	n
	-127.6190476	difference (USA - Europe)
	165.2219287	standard error of difference
	0	hypothesized difference
	-0.77	z
	.4399	p-value (two-tailed)

The p-value is 0.4399.

Since the p-value (0.4399) is greater than the significance level (0.01), we fail to reject the null hypothesis.

Therefore, we can conclude that the average length of the major rivers in the United States is the same as the average length of the major rivers in Europe.