Question 1 Question 2 Question 3 Question 4 Question 5 Question 6. Math 144 tutorial April, 2010

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1 Math 144 tutorial 7 12 April, 2010

2 1. Let us define S 2 = 1 n n i=1 (X i X) 2. Show that E(S 2 ) = n 1 n σ2.

3 1. Let us define S 2 = 1 n n i=1 (X i X) 2. Show that E(S 2 ) = n 1 n σ2.

4 1. Let us define S 2 = 1 n n i=1 (X i X) 2. Show that E(S 2 ) = n 1 n σ2. This result also indicate that S 2 is a biased estimator of the σ 2.

5 2. If X is a binomial random variable, show that (a). ˆp = X/n is an unbiased estimator of p; (b). p = X+ n/2 n+ n is a biased estimator of p; (c). Show that the estimator p becomes unbiased as n.

6 2. If X is a binomial random variable, show that (a). ˆp = X/n is an unbiased estimator of p; (b). p = X+ n/2 n+ n is a biased estimator of p; (c). Show that the estimator p becomes unbiased as n. Solution: (a). Since E[ˆp] = E[X]/n = np/n = p, this shows that ˆp is an unbiased estimator of p. (b). E[p ] = E[X]+ n/2 n+ n = np+ n/2 n+ n p. (c). np + n/2 lim n n + n p = lim n 1 + n/n + lim 1 n 2(1 + n) = p.

7 3. An electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed with a standard deviation of 40 hours. If a sample of 30 bulbs has an average life of 780 hours. (a). Find a 96% confidence interval for population mean of all bulbs produced by this firm; (b). How large a sample is needed if we wish to be 96% confident that our sample mean will be within 10 hours of the true mean?

8 3. An electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed with a standard deviation of 40 hours. If a sample of 30 bulbs has an average life of 780 hours. (a). Find a 96% confidence interval for population mean of all bulbs produced by this firm; (b). How large a sample is needed if we wish to be 96% confident that our sample mean will be within 10 hours of the true mean? Solution: The sample mean x is 780, and the standard deviation of population is 40, using the Normal table, we can get z 0.02 = Hence the 96% confidence interval for µ is [ x z 0.02 ( σ 30 ), x + z 0.02 ( σ 30 )] = [ , ]. (b). We can be 96% confident that this error will less than z α/2 ( s n ), i.e, n = ( z e ) 2 = Therefore, a sample with size 68 is needed if we wish be 96% confident that our sample mean will be within 10 hours of the true mean.

9 4. Many cardiac patients wear implanted pacemakers to control their hear-beat. A plastic connector module mounts on the top of the pacemaker. Assuming a standard deviation of and an approximate normal distribution, (a). Find a 95% confidence interval for the mean of all connector modules made by a certain manufacturing company. A random sample of 75 modules has an average of inch. (b). How large a sample is needed if we wish to be 95% confident that our sample mean will be within inch of true mean?

10 (b). n = ( z ) 2 = That is, a sample with size 35 is needed if we wish to be 95% confident that our sample mean will be within inch of true mean. 4. Many cardiac patients wear implanted pacemakers to control their hear-beat. A plastic connector module mounts on the top of the pacemaker. Assuming a standard deviation of and an approximate normal distribution, (a). Find a 95% confidence interval for the mean of all connector modules made by a certain manufacturing company. A random sample of 75 modules has an average of inch. (b). How large a sample is needed if we wish to be 95% confident that our sample mean will be within inch of true mean? Solution (a). From the problem, we obtain n = 75, x = 0.310, σ = , hence the 95% confidence interval for the mean of all modules is: [0.310 z ( ), z ( )] = [ , ]

11 5.The heights of a random sample of 50 college students showed a mean of centimeters and a standard deviation of 6.9 centimeters. (a). Construct a 98% confidence interval for the mean height of all college students. (b). What can we assert with 98% confidence about the possible size of our error it we estimate the mean height of all college students to be centimeters?

12 5.The heights of a random sample of 50 college students showed a mean of centimeters and a standard deviation of 6.9 centimeters. (a). Construct a 98% confidence interval for the mean height of all college students. (b). What can we assert with 98% confidence about the possible size of our error it we estimate the mean height of all college students to be centimeters? Solution: Because standard deviation of the population is unknown, so we should use the t distribution: (a). Because s = 6.9, x = 174.5, n = 50, so 98% confidence interval for the mean height of all college students is: [174.5.t 0.01,49 ( ), t 0.01,49 ( )] = [ , ]. (b). We can be 98% confident that this error will less than t α/2 ( s n ), i.e, e t 0.01,49 ( ) =

13 6. A random sample of 100 automobile owners shows that, in the state of Virginia, an automobile is driven on the average 23,500 kilometers per year with a standard deviation of 3900 kilometers. Assume the distribution of measurements to be approximately normal. (a). Construct a 99% confidence interval for the average number of kilometers an automobile is driven annually in Virginia. (b). What can we assert with 99% confidence about the possible size of our error if we estimate the average number of kilometers driven by car owners in Virginia to be 23,500 kilometers per year?

14 6. A random sample of 100 automobile owners shows that, in the state of Virginia, an automobile is driven on the average 23,500 kilometers per year with a standard deviation of 3900 kilometers. Assume the distribution of measurements to be approximately normal. (a). Construct a 99% confidence interval for the average number of kilometers an automobile is driven annually in Virginia. (b). What can we assert with 99% confidence about the possible size of our error if we estimate the average number of kilometers driven by car owners in Virginia to be 23,500 kilometers per year? Solution: (a). We use the t distribution, x = 23500, s = 3900, n = 100, then the 99% confidence interval for population mean µ is: [23500 t 0.005,99 ( ), t 0.005,99 ( )] = [ , ]. (b). e t 0.005,99 ( ) =

The problems of first TA class of Math144

The problems of first TA class of Math144 T eaching Assistant : Liu Zhi Date : F riday, F eb 2, 27 Q1. According to the journal Chemical Engineering, an important property of a fiber is its water absorbency.