8-1-1 CONFIDENCE INTERVALS, Assignments of Mathematics

Download 8-1-1 CONFIDENCE INTERVALS and more Assignments Mathematics in PDF only on Docsity! 8.1.1 Confidence Intervals 1 / 7 1. Find the confidence interval: (a) for the true mean height of Amer- ican women, given a sample mean of x¯=64 inches, and margin of error 2.4 inches. (b) for the true proportion of voters who support an issue, given a sample (a) The point estimate is the sample mean x¯=64 inches, and the margin of error is 2.4 inch- es. The confidence interval is: (64 2.4,64+2.4)(61.6,66.4) (b) The point estimate is the sample proportion p =0.45, proportion of p =0.45, and margin of and the margin of error is error 0.04. 2. A researcher is trying to estimate the population mean for a certain set of data. The sample mean is x¯=45, and the margin of error for the mean is 9. Find the corresponding confidence in- terval for the mean. 3. In a recent questionnaire about home ownership, a random sample of 545 homeowners were asked about fixed and adjustable rate mortgages, and 268 reported that they have an ad- justable rate mortgage. What value of z should be used to cal- culate a confidence interval with a 95% confidence level? z0.10 z0.05 z0.025 z0.01 z0.005 1.282 1.645 1.960 2.326 2.576 4. The lengths, in inches, of adult corn snakes are normally distributed with a population standard deviation of 6 inches and an unknown population mean. If a random sample of 18 snakes is taken and results in a sample mean 0.04. The confidence interval is: (0.45 0.04,0.45+0.04)(0.41,0.49) We are given the sample mean x¯=45, and a margin of error of 9. So the confidence interval esti- mate, at the 99.7% confidence lev- el, is (45 9,45+9), which simplifies to the confidence interval (36,54). 2.326. We can use the formula to find the margin of error: ME=(z ±/2)(nÃ)/ We know that Ã=6and n=18. We are also given that the confidence level 8.1.1 Confidence Intervals 2 / 7 of 61 inches, find the margin of error (ME) of the confidence interval with a 90% confidence level. Round your an- swer to three decimal places. z0.10 z0.05 z0.025 z0.01 z0.005 1.282 1.645 1.960 2.326 2.576 5. Suppose heights, in inches, of orang- utans are normally distributed and have a known population standard de- viation of 4 inches and an unknown population mean. A random sample of 16 orangutans is taken and gives a sample mean of 56 inches. Find the confidence interval for the population mean with a 95% confidence level. z0.10 z0.05 z0.025 z0.01 z0.005 1.282 1.645 1.960 2.326 2.576 Round the final answer to two decimal places. (CL) is 90%, or 0.9. So, we can calculate alpha (±). ±=1C=L=1 =0.9 =0.1 Since ±=0.1,we know that ±2 =0.12 =0.05 The value of z0.05 is 1.645. Now we can substitute the values into the formula to find the margin of error. ME =(z ±/2)(nÃ)/ =(1.645)(6/ 18) H(1.645)(1.414) H2.326 So, the margin of error (ME) is 2.326. (54.04 , 57.96) use the Confidence interval equa- tion. The point estimate is the sample mean, x¯, and the margin of error is margin of error = (z ±/2)(nÃ)/ Substituting the given values Ã=4, n=16, and z ±/2=1.96for a confi- dence level of 95%, we have margin of error=(1.96)(4/ 16) H(1.96)(1.0) H1.96 With x=56 and a margin of error of 1.96, the confidence interval is (56 1.96 , 56 + 1.96) (54.04,57.96). So we estimate with 95% con- fidence that the true population mean is between 54.04 and 57.96 inches. 8.1.1 Confidence Intervals 5 / 7 amusement parks is taken and results n=22, and z ±/2 =2.576for a confi- in a sample mean of 61 dollars. Find the margin of error for a 99% confidence interval for the population mean. z0.10 z0.05 z0.025 z0.01 z0.005 1.282 1.645 1.960 2.326 2.576 12. Suppose heights of seasonal pine saplings are normally distributed and have a known population standard de- viation of 17 millimeters and an un- known population mean. A random sample of 15 saplings is taken and gives a sample mean of 308 millime- ters. Find the confidence interval for the population mean with a 90% confi- dence level. z0.10 z0.05 z0.025 z0.01 z0.005 1.282 1.645 1.960 2.326 2.576 13. A bank offers auto loans to qualified customers. The amount of the loans are normally distributed and have a known population standard deviation of 4 thousand dollars and an unknown population mean. A random sample of 22 loans is taken and gives a sample mean of 42 thousand dollars. Find the margin of error for the confi- dence interval for the population mean with a 90% confidence level. dence level of 99%, we have mar- gin of error =(2.576)(2.5/ 22) H(2.576)(0.53) H1.37 Ã=17 n=15 z ±/2 =1.645 confidence level of 90%, we find the margin of error = (1.645)(17/ 15) H (1.645)(4.389) H7.22 x¯=308 subtract M.O.E = 7.22, the confidence interval: (308 7.22 , 308 + 7.22) = (300.78,315.22). So we estimate with 90% con- fidence that the true population mean is between 300.78 and 315.22 millimeters. 1.40 Ã=4n, =22, and z ±/2 = 1.645for a confidence level of 90%, we have margin of error: =(1.645)(4/ 22) H(1.645)(0.852) H1.40 8.1.1 Confidence Intervals 6 / 7 z0.10 z0.05 z0.025 z0.01 z0.005 1.282 1.645 1.960 2.326 2.576 14. The weekly salaries of sociologists in the United States are normally dis- tributed and have a known population standard deviation of 425 dollars and an unknown population mean. A ran- dom sample of 22 sociologists is tak- en and gives a sample mean of 1520 dollars. Find the margin of error for the confi- dence interval for the population mean with a 98% confidence level. z0.10 z0.05 z0.025 z0.01 z0.005 1.282 1.645 1.960 2.326 2.576 15. Suppose the finishing times for cy- clists in a race are normally distrib- uted. If the population standard devia- tion is 16 minutes, what minimum sam- ple size is needed to be 95% confident that the sample mean is within 5 min- utes of the true population mean? z0.10 z0.05 z0.025 z0.01 z0.005 1.282 1.645 1.960 2.326 2.576 16. Suppose the germination periods, in days, for grass seed are normally dis- tributed. If the population standard de- viation is 3 days, what minimum sam- ple size is needed to be 90% confident that the sample mean is within 1 day of the true population mean? 210.76 values Ã=425n, =22, and z±2=2.326 for a confidence level of 98%, we have margin of error =(2.326)(425/ 22) H(2.326)(90.61) H210.76 formula used z =z ±/2 = z0.025 =1.96 because the confidence level is95%. Ã= 16and ME = 5. Therefore, n = z^2Ã^2M/ E^2 =(1.96)^2(16)^2/5^2 H39.34 Use n= 40 to ensure that the sam- ple size is large enough. formula used z =z ±/2 = z0.05 =1.645 because the confidence level is90%. Ã= 3and ME = 1. Therefore, n = z^2Ã^2M/ E^2 =(1.645)^2(3)^2/1^2 H 8.1.1 Confidence Intervals 7 / 7 z0.10 z0.05 z0.025 z0.01 z0.005 1.282 1.645 1.960 2.326 2.576 17. The population standard deviation for the typing speeds for secretaries is 4 words per minute. If we want to be 90% confident that the sample mean is within 1 word per minute of the true population mean, what is the minimum sample size that should be taken? Use n= 40 to ensure that the sam- ple size is large enough. H43.30