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Unformatted text preview: Activity 2Solutions 1. A random sample of size n = 30 is taken from a population with mean μ and standard deviation σ. Which statement is generally correct? (4 pts) a. X is an estimate of μ; s is an estimate of σ. b. μ is an estimate of X ; σ is an estimate of s. c. μ is an estimate of X ; s is an estimate of the standard deviation of the sample mean. d. X is an estimate of μ; s is an estimate of the standard deviation of the sample mean. e. X is an estimate of μ; s is the standard error of the sample mean. 2. Suppose we draw a random sample from a population which is normally distributed with mean 20 and s.d. 5. Assuming the sample points are drawn independently, which of the values can the sample size be equal to, so that the sample mean follows a normal distribution? ( 4 pts) a. b. c. d. e. 3 20 40 All of the above None of the above 3. As the sample size increases, the standard error of the sample mean ____________(select the best response): ( 4 pts) a. b. c. d. increases decreases stays the same None of the above. distribution. Let X and Y be independent. 4. Let X be a N(0,1) random variable and Y have a What is the distribution of ? ( 4 pts) √ a. t-distribution with 9 degrees of freedom. b. t-distribution with 8 degrees of freedom. c. Normal distribution with mean 0 and variance 1. d. None of the above . 5. If Z=(X+4)/3 follows a N(0,1) distribution. What is the distribution of X? ( 4 pts) a. Normal distribution with mean 4 and variance 3. b. Normal distribution with mean 4 and standard deviation 3. c. Normal distribution with mean -4 and variance 3. d. Normal distribution with mean -4 and standard deviation 3. e. None of the above Show your Work to get full credit: 6. Suppose X is a continuous random variable following a uniform distribution over the interval (a,b). Suppose E(X) =10 and Var(X)=3. What is the probability that X lies between 11 and 15? ( 8 pts) E(X)=(a+b)/2=10, Var(X)=(b-a)2/12=3. Hence a+b=20 and b-a=±6 Solving the two equations, a=7 and b=13 (note a has to be less than b) Hence f(x) = 1 6 0
15 for 7 x 13 o.w.
13 15 1 So, P(11<X<15)= f ( x)dx = dx 0dx =1/3. 6 11 11 13 7. Suppose that samples of size n=25 are selected at random from a normal population with mean 100 and standard deviation 10. a. What is the probability that the sample mean falls in the interval from 1.0 ? ( 8 pts) Note , the sample is from normal population. So even if n=25, distribution. P( 1.8 =P(-1.8<
1.0) 1.0 1.8 to follows a N(0,1) ) =P(-1.8<Z<1.0) =P(Z<1.0)-P(Z<-1.8) =0.8413-0.0359 =0.8054. b. How large must the random sample be so that the standard error of the sample mean is 2.5? ( 6 pts) Standar error of = . Here
√ √ 2.5, i.e., √ 2.5. So n=
√ 16. . had been (Here population standard deviation is known, so standard error is unknown, then would be estimated by s, the sample s.d.. In that case the estimated standard error would have been )
√ 8. For a t-distribution with 23 degrees of freedom, find k such that: (6 pts) < k)= 0.965. a. P( 2.069 From Table A.4 we note that 2.069 corresponds to t0.025 when v = 23. Therefore, −t0.025 = −2.069 which means that the total area under the curve to the left of t = k is 0.025 + 0.965 = 0.990. Hence, k = t0.01 = 2.500. = 0.10. Hence, k = t0.10 = 1.319. b. P(-k < T < k)=0.90. t0.05 = 1.714 for 23 degrees of freedom. So k=1.714. 9. A manufacturing firm claims that batteries used in their games last 30 hours on the average. Suppose you draw a sample of 16 batteries to verify this claim. The sample mean is equal to 27.5 hours with a standard deviation of 5 hours. You can assume that distribution of battery lives is X approximately normal. Suppose the claim is taken to be valid if the calculated statistic S/ n follows in the central 95% of its distribution. What distribution does the statistic follow? Does the claim seem to be valid? (8 points) Under the given conditions, you can assume that the original population is normal. X Here the statistic follows a t-distribution with 15 degrees of freedom (Recall for small S/ n X n, if standard deviation of population is unknown, then does NOT follow a N(0,1) S/ n distribution even though the original population is normal) . , s=5 yrs. According to the claim, population mean (i.e. the average Here n=16, ̅ 27.5 lifetime of all the batteries) is 30 hrs. So, μ=30 yrs. So calculated statistic= and . . ; ; . /√ 2. The central 95% of the t-distribution is between . ; . From the table . ; 2.131>2. So the calculated statistic lies between and . ; . So based on the sample, the manufacturing claim seems to be valid. ...
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This note was uploaded on 01/25/2011 for the course MATH 380 taught by Professor Staff during the Fall '08 term at UNL.
- Fall '08
- Standard Deviation