Question

Consider two independent random samples of sizes n₁ = 14 and n₂ = 10, taken from two normally distributed populations. The sample standard deviations are calculated to be s₁= 2.32 and s₂ = 6.74, and the sample means are x¯1=-10.1and x¯2=-2.19, respectively. Using this information, test the null hypothesis H0:μ1=μ2against the one-sided alternative HA:μ1<μ2, using the Welch Approximate t Procedure (i.e. assuming that the population variances are not equal).

a) Calculate the value for the t test statistic.

Round your response to at least 3 decimal places.

b)  Using the Welch-Satterthwaite approximate degrees of freedom of 10.535573, the p-value is within which one of the following ranges?

		p-value > 0.10
		0.05 < p-value < 0.10
		0.01 < p-value < 0.05
		0.005 < p-value < 0.01
		p-value < 0.005

c) What is the most appropriate conclusion that can be made, at the 1% level of significance?

		There is sufficient evidence to reject the null hypothesis in favour of the alternative, that the mean of Population 1 is less than that of Population 2.
		There is insufficient evidence to reject the null hypothesis, and therefore no significant evidence that the two population means are different.
		We can be completely certain that the mean of Population 1 is less than the mean of Population 2, as the p-value is very small.
		We can be completely certain the that means of the two populations are equal, as the p-value is very large.
		The results of the hypothesis test are invalid, since the assumptions of the Welch approximate t procedure were not met.

Answer 1

The statistical software output for this problem is:

Hence,

a) t = -3.563

b) P-value < 0.005

c) There is sufficient evidence to reject the null hypothesis in favour of the alternative, that the mean of Population 1 is less than that of Population 2.