10.1 Mean sd Known

# 86 and balls averaged 258 yards based nn estimate 1 2

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Unformatted text preview: heir competitor, Ace, Inc. σd = n + n standard In a test of driving distance using a We are 95% confident = 2.622 error mechanical driving device, a sample of Par tgolf balls was compared in a hat the difference with 2 2 margin of za 2 snPar + snAce = 5.139 sample of Ace golf balls. In a sample of Par Ace mean driving distances error 120 balls, the Par balls averaged 275 b and in a the 2 golf balls σ2 σ 2 yardsetweensample of 80 balls, the interval d ±z 2 1 + 2 α Aceis between 11.86 and balls averaged 258 yards. Based nn estimate 1 2 Given that 0 is not within the on data from previous driving distance 22.14 yards. interval, we can be 95% confident tests, the two population standard deviations are known with σPar = 15 that there is a difference between the mean driving distances of the yards and σAce = 20 yards. golf balls. 2 2 Par Ace Par Ace Hypothesis Tests Hypothesis Tests about the Difference between Two Population Means (σ1 and σ2 known) difference between 2 population means: D = μ1 – μ2 to ◦ test hypotheses about D, d x− 2 =1 x take independent simple random samples from 2 populations d x− =x 1 ◦ use sampling distribution of Hypothesis Tests ◦ 2 if both populations follow normal distribution or if n1 and n2 are large enough for Central Limit Theorem to apply, then will have a normal distribution decision rules based on either p-value or critical value approach calculate test statistic and decide based on decision rule Hypothesis Tes...
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