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Stat 145: Exam 4 Review Answers 1. (a) The sample proportion is: ˆ p = 459 850 = 0 . 54 (b) A 95% confidence interval for p is: 0 . 54 ± 1 . 96 0 . 54(1 - 0 . 54) 850 = 0 . 54 ± 0 . 0335 = (0 . 5065 , 0 . 5735) (c) The sample size should be: n = 1 . 96 0 . 03 2 (0 . 54)(1 - 0 . 54) = 1060 . 2816 Rounding up to the next higher whole number, use a sample size of n = 1061. 2. (a) The sample mean is: ¯ x = 58 + 47 + 57 + 55 4 = 54 . 25 and the sample standard deviation is: s = (58 - 54 . 25) 2 + (47 - 54 . 25) 2 + (57 - 54 . 25) 2 + (55 - 54 . 25) 2 4 - 1 = 74 . 75 3 = 4 . 99 (b) The hypotheses are: H 0 : μ = 51 H A : μ > 51 The value of the test statistic is: t = 54 . 25 - 51 4 . 99 / 4 . = 1 . 30

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There are 4 - 1 = 3 degrees of freedom, so the P -value is: P = P ( T 1 . 30) between . 10 and . 15 We fail to reject the null hypothesis and conclude that the current mean retail price for bananas is not greater than 51 cents per pound. 3. (a) The stemplot appears below. 2 79 3 45 4 26788 5 36 6 144 7 1 (b) Yes, because the distribution is roughly Normal with no apparent outliers. 4. (a) The degrees of freedom are equal to 31. This value is not on the table, so use the row with 30 df. A 90% confidence interval for μ 1 - μ 2 is: (383 . 4 - 349 . 6) ± 1 . 697 37 . 2 2 32 + 34 . 6 2 40 .
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