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As with the confidence intervals, we will be assuming the two population standard deviations are equa but unknown. Thus, we will need to calculate the pooled standard deviation to use in our test statistic. Dur typical hypothesis test would be of the form H:44-42 = 8vs. H:4,-4, 78This is a two side test because the alternative is not equal. (H:14-4, <8, or H:44 - H2>, are also possible. Our test statistic is I X - X₂-8 1 1 wher (n, - 1)s} +(n, -1)s n+n₂ - 2 Spam S + Example: A sample of 10 pieces of data is taken from the first population. The mean of the data is 12.5. A sample of 12 pieces of data is taken from the second population. The mean of the data is 12.42. Perform a .05 level test to determine if the population means differ. Assume that S =.055.

Answer 1

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( Here , Critical value calculate form t-distribution table.)

classmate Date Page paged solution . given that A sample of 10 pieces of data is taken from the popwatian and mean of the data is 12.5' and A sample of 12 pieces of the data is taken from the second population population the the mean mean of data of data is 12.42 fevel of significance ax=0.05 from above information for ist population n=10 = 12.5 for and populattan - n = 12 = 12.42. we want to test - Ha el 12 against Hi e, & 12. Test statistici. Here sample size h=10 M2 = 12 230 - We use t- test statistic. - ntnm (under Ho) spt that he = Sp= 0.055 is given ! 12.5-12-42 0.055 10 12 | t - O. 7967 coltical value = tnoth2-2 & t 10-12-2, 0.025 a tzo, 0.025 I tnt02-24/= 2.086 ] Decision sele; if 1t17 tnit na-2.as then we heiecte otherwise Accept Ho Here It=0.7967 tnith₂-2,5/2 = 2.0 We accept Ho at 5 % level of. cance TH conjusion There is no difference between populations means
Conclusion : There is no difference between population means i.e. mean of population first is equal to mean of population second .