Ans - 6 – Intro

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Name: ________________________________ Statistics | Rutgers-Newark | Saleh 6 – Intro. to Inference Show your work. 1. Suppose a population that has mean 90 and standard deviation 20. How many observations, n, should there so that the standard deviation of the sampling distribution of 𝑥̅ is 1? ???????? ??𝑣𝑖??𝑖?? ?? ?????? = 𝜎 √? = 1 = 20 √? ? = 20 2 = 400 2. According to the CDC, the average height of American men 20 years old and up is 69 inches. Assume the distribution of height is Normally distributed with a standard deviation of 3 inches. a. What is the probability of randomly selecting a man taller than 72 inches? First, find Z-score: 72 − 69 3 = 1 Probability (area to the left of z) using Table A → 0.8413 Need to find probability above 72: 1 – 0.8413 = 0.1587 or 15.87% b. What is the probability of randomly selecting three men with a sample mean height greater than 72 inches? 72 − 69 3/√3 = 1.73 Denominator = sigma/sqrt(n) Probability (area to the left of z) using Table A → 0.9582 Need to find probability above 72: 1 – 0.9582 = 0.0418 or 4.18% c. What is the probability of randomly selecting three men all with heights greater than 72 inches? 0.1587 3 = 0.00399 3. Suppose we want a 90% confidence interval for the average amount of money spent on a sofa with a margin of error of $50. The standard deviation is $100. What would be the required number of observations? With a 90% CI, the corresponding critical z value, z* = 1.645 m = 𝑧 ∗ ( 𝜎 √𝑛 ) = $50 = 1.645 ( $50 √𝑛 ) ? = 10.76 ?? 11 ?????𝑣??𝑖???
4. Suppose a population is Normally distributed with σ = 4 A random sample of 25 observations is drawn from the population and we find the sample mean of these observations is 𝑥̅ = 12. We test the null hypothesis H 0 : μ = 10 and the alternative Ha: μ > 10. a. What is the value of the test statistic z? 12 − 10 4/√25 = 2 0.80 = 2.5 b. What is the P-value? Use Table A → 0.9938 (area/probability to the left of z) P-value is the probability of observing a value of Z at least as extreme as the observed z=2.5 P(Z ≥ 2.5) = 1- 0.9938 = 0.0062 c. Would you reject or fail to reject the null at the 5% significance level? Why? Reject null as p > α d. Would you reject or fail to reject the null at the 1% significance level? Reject null as p > α 5. Suppose test scores can range from 0 to 200 and follow an (approximately) Normal distribution with mean 115 and standard deviation σ = 25. H 0 : μ = 120 and the alternative Ha: μ≠ 120. A class of 30 students had a mean score of 117. a. What is the value of the test statistic z? 117 − 120 25/√30 = 0.657 b. What is the P-value? This is two-sided test. We need to find the area of one side/tail and then double that value. Use Table A to find the probability of +z or -z: If using -z (i.e. -0.66) → area = 0.2546 P-value = 2(0.2546) = 0.5092 If using +0.66 → area = 0.7454 (area/probability to the left of z) Right tail = 1 – 0.7454 = 0.2546 2(0.2546) = 0.5092 c. Would you reject or fail to reject the null at the 5% significance level? Why? Fail to reject null as p > α
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