. 13. The maximum error (margin of error) of the estimate for μ (based on known σ) is:

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1 Statistics Sample Exam 3 Solution Chapters 6 & 7: Normal Probability Distributions & Estimates 1. What percent of normally distributed data value lie within 2 standard deviations to either side of the mean? About 95% Or: Answer: P (-2 < z < 2) = P (2) P (-2) = = or 95.44% 2. For a continuous random variable, the probability of a single value of x is always 0 3. The parameters of normal distribution are and. 4. What are the values of mean and the standard deviation for the standard normal distribution? 0 and 1 5. Find the value of z for a standard normal distribution, such that the area in the left tail is 0.05 z = Find the value of z for a standard normal distribution, such that the area in the right tail is = 0.9 (cumulative area from the left) 1.28 < z < 1.29 z = Find the following probability: P(z>-0.82) The z-value for μ of normal distribution curve is always 0 9. Usually the normal distribution is used as an approximation to binomial distribution when np >5, and nq True or False: The confidence interval estimate of the population mean is constructed around the sample mean. True 11. Estimation means assigning values to a population parameter based on the value of a sample 12. The confidence level is denoted by (1 )100%. 13. The maximum error (margin of error) of the estimate for μ (based on known σ) is: E Z /2 n 14. The parameter(s) of t distribution is (are) degrees of freedom: df 15. What criteria are required to apply the t distribution to make a confidence interval for μ? a) The sample must be a simple random sample b) Either the sample is from a normally distributed population or n 30

2 c) Population Standard Deviation, σ is unknown 16. Find the values for t for a t-distribution with sample size of 20 and a confidence level of 95%. degrees of freedom: df = n 1 = 19, t Find the values for t for a t-distribution with sample size of 30 and a confidence level of 90%. degrees of freedom: n 1 = 29, t Find the critical chi-square value for 20 degrees of freedom when the area to the left is 0.01 = = Find the critical chi-square value for n=30 when the area to the right is 0.9 degrees of freedom: 29 = = Three out of eight drinkers acquire the habit by age 20. If 300 drinkers are randomly selected, using normal approximation to the binomial distribution, find the probability that: Given: n = 300, p = 3/8 (0.375), q = (1-p) a. Exactly 100 acquire the habit by age 20 P(x = 100) = P( 99.5 < x < 100.5) z = x μ / σ, μ = np = 300 x = 112.5, σ 2 = npq = σ = z 1 = 99.5 (300 x 0.375) / (300 x x 0.625) z 1 = / = z 2 = / = -12 / = P( < z < ) = = b. At least 125 acquired the habit by age 20 P(x 125) = P(x > 124.5) z = ( ) / = P(z > ) = = c. At most 165 acquired the habit by age 20 P(x 165) = P(x < 165.5) z = ( ) / = above 3.5 in the Z-table P(z < ) = IQ scores are normally distributed with a mean of 100 and a standard deviation of 15.

3 a. What percentage of people has an IQ score above 120? Given: normal distribution, μ = 100 σ = 15 P(x > 120) z = x μ /σ = ( ) / 15 = 1.33 = P(z > 1.33) = = , 9.18% b. What is the maximum IQ score for the bottom 30%? need to find x when area under the curve corresponds to 0.3 x = zσ + μ < z < x = ( x 15) = x = maximum IQ score for the bottom 30% is Men have head breaths that are normally distributed with a mean of 6.0 in and a standard deviation of 1.0 in. a. If a man is randomly selected, find the probability that his head breadth is less than 6.2 in. given: normal distribution, n = 1, : μ = 6.0 in., σ = 1.0 in. P( x < 6.2) z = x μ / σ = z = ( )/1 = 0.2 Answer: P (x < 6.2) = P (z < 0.2) = If 100 men are randomly selected, find the probability that their head breadth mean is less than 6.2 in. Answer: Given: n = in / n = 1.0/ = 0.1 x z = ( )/0.1 = 2 P ( x < 6.2) = P (z < 2) = x 23. A statistician is interested in estimating at a 95% confidence level the mean number of houses sold per month by all real state agents in a large city. It is known that the population standard deviation is 1.9. a. How large a sample should be taken so that the estimate is within 0.65 of the population mean? n = [(zα/2 σ)/e] 2 = [(1.96)(1.9)/0.65] 2 = (Rounded up) b. Find this 95% confidence interval, if the mean for such sample size is 3.4 x = 3.4, z α/2 = 1.96, E = 0.65 x E < μ < x + E & < μ < 4.05

4 24. A random sample of 20 female members of a health club showed that they spend on average 4.5 hours per week doing physical exercise with standard deviation of 0.75 hours (assume the time spent on exercise for all female members is approximately normally distributed) Given: n = 20, x = 4.5 hr & s = 0.75 hr a. What is the value of the point estimate of the population mean? point estimate of μ, the population mean is: x = 4.5 hours per week b. Construct the 98% confidence for the mean time spent on physical exercise of all such females. degree of freedom: df = 20 1 = 19 CI = 0.98 α = 0.02 t α/2 = E = t α/2 (s/ ) = (2.539)(0.75/ ) = hours 98% CI: x E < μ < x + E < μ < Out of 500 randomly selected adults, 300 said they were in favor of the death penalty for a person convicted of murder. Given: n = 500, x = 300 a. What is the value of point estimate of the population proportion? p = x/n = 3/5 = 0.6 b. What is the 95% confidence interval for the population proportion? CI = 0.95 α = 0.05 z α/2 = 1.96, q = = 0.4 E = z α/2 = (1.96) = E = 1.96 x ( 0.6 x 0.4 / 500) = p - E < p < p +E < p < How large a sample is needed to be 99% confident that the margin of error is 0.025, for percentage of golfers that are left-handed, if we a. Assume that 15% of the golfers are left-handed, based on a previous study. E = 0.025, CI = 0.99, z α/2 = 2.575, p = 0.15 n = (z α/2 ) 2 p q / E 2 n = [(2.575) 2 x 0.15 x 0.85] / n = ; but rounds up to 1353 Assume that we have no prior information suggesting a possible value for the sample proportion. assume p = q = 0.5 n = = [(2.575) 2 (0.25)]/(0.025) 2 = (rounded UP to 2653) 27. Given: n 30, x 80.5, s 4.6 (assume normal population)

5 a. Find 98% confidence interval for population variance df = n 1 = = X 2 R X 2 L = b. Find 98% confidence interval for population standard deviation square root value of part a

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