In: Math
Questions 1-5 are based on the following Let X be a normally distributed random variable. A random sample of size n = 9 yields the following data: x 98 93 61 75 58 75 95 77 70 1 The variance of the sample is, a 210.75 b 201.21 c 191.67 d 187.33 2 The variance of x̅ is, a 62.229 b 74.083 c 23.417 d 34.572 3 You want to build a 95% confidence interval for the population mean. The margin of error (MOE) for the interval is, a 12.836 b 11.159 c 10.705 d 9.485 4 The 95% interval estimate is, a 66.841 89.159 b 67.064 88.936 c 68.515 87.485 d 69.464 86.536 5 You want to build an interval estimate that would capture the population mean within ±3 from the sample mean 95% of the time. What is the minimum sample size to yield such an interval? Round the standard deviation from the above sample data to the nearest integer for the planning value. a 104 b 97 c 89 d 81
X | ∑x² |
98 | 9604 |
93 | 8649 |
61 | 3721 |
75 | 5625 |
58 | 3364 |
75 | 5625 |
95 | 9025 |
77 | 5929 |
70 | 4900 |
∑x = | ∑x² = |
702 | 56442 |
Mean , x̅ = Ʃx/n = 702/9 = 78
1 The variance of the sample :
Variance, s² = (Ʃx² - (Ʃx)²/n)/(n-1) = (56442-(702)²/9)/(9-1) = 210.75
Answer a.
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2 The variance of x̅ = s²/n = 210.75/9 = 23.417
Answer c.
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3. At α = 0.05 and df = n-1 = 8, two tailed critical value, t-crit = T.INV.2T(0.05, 8) = 2.306
Margin of error , E = t*√(s²/n) = 2.306*√23.417= 11.159
Answer b.
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4 The 95% interval estimate :
Lower Bound = x̅ - E = 78 - 11.1589 = 66.841
Upper Bound = x̅ + E = 78 + 11.1589 = 89.159
Answer a.
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5. Standard deviation, s = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(56442-(702)²/9)/(9-1)] = 14.517 = 15
Margin of error, E = 3
Confidence interval, CL = 0.95
Significance level, α = 1-CL = 0.05
Critical value, z = NORM.S.INV(0.05/2) = 1.9600
Sample size, n = (z * σ / E)² = (1.96 * 15 / 3)² = 96.04 = 97
Answer b.