University:
McMaster University
Course:
STATS 1LL3 | Introduction to Probability and Statistics
Academic year:

2023
Views:

457

Pages:

3
Author:

Sabrina S.

Comparing two independent sample means (variance estimated from the sample) When we have small samples from two populations and want to compare their means, the procedure is a little more complex than one might expect. z = -- (۰) (٣٠-٣٠) Na you might anticipate that the test statistic would be S/S / But, unfortunately, statisticians can show that this quantity does not have the t-distribution In order to find a test statistic which does have the t-distribution, an additional assumption needs to be made. The assumption is that the population variances in the two populations from which the samples were drawn are equal. Example (highlights difference between comparing means when variance is known and when variance is estimated from sample) Two varieties of wheat being tested in a country. Twelve test plots are given identical preparatory treatment. Sir plats are sown with variety 1 and other siy with variety 2 Scientists hope to determine significant difference, 5% level. Variety 1. 1-5 12 2.0 23 tons per plot Variety 2. 16 11 1.1 2-3 tons per plat One of the plots in variety 2 was accidently given extra fertilizer Results discardlead X 160 S1 0-42 6 1.१० S 0-27 We follow the standard hypothesis testing procedure Ho - N 2.) H1N1 (a two-tailed test) 3.) significance level 5% 4. $ s.) Before we find the rejection region we need to know the degrees of freedom". The procedure is rather different from that in the test when the population variances were known. Instead of working L with the individual variances on and as we assume that both populations have the same variance and we pool the two individual sample variances 2 2 S₁ and S. to form a joint estimate of the variance, s. This assumption of equal variances is required by the mathematical theory underlying the t-distribution. the general formula for pooled variance s (n-1)s. + (n-1) S 2 S2 therefore, s5x(0.42)² + 4x (027) 6+5-2 0.13 Standard deviation 1013 0-361 How many degrees of freedom does s have? ++ १ Thus we use the t-distribution with 9 df and because we have a two-sided alternative and a 5% significance level we need the value of 0.025 from the table is 2.262 observed ty value 15 less than -2.262 So we reject Ho if the or greater than 2.262. two-sample t-test 1 - 2 - (μ- με) thi+n2-2- and X2 are the sample means Na is determined by the null hypothesis N1 S is pooled sample standard deviation. nt and ne are the sample sizes 1.60 1.90 (0) t 6+5-2 0.3611+5 691372 6.) Becouse -1.372 does not lie in the rejection region we conclude that the difference between the varieties is not significant.