STP231 HW12
requirements for constructing a confidence interval for estimating a population mean with sigma known?
-The sample measures a quantitative value. -Either the population is normally distributed or ngreater than30, or both. -The sample is a simple random sample.
Properties of student t distribution
-The standard deviation of the Student t distribution is s=1. -The Student t distribution has the same general symmetric bell shape as the standard normal distribution, but it reflects the greater variability that is expected with small samples. -The Student t distribution is different for different sample sizes.
Which of the following is NOT an equivalent expression for the confidence interval given by 161.7 less than mu less than 189.5?
161.7 plus or minus 27.8
Which of the following is NOT a requirement for constructing a confidence interval for estimating a population mean with sigma known?
A confidence level of 95% is not a requirement for constructing a confidence interval to estimate the mean with sigma known. The confidence level may vary.
A 95% confidence interval of 17.6 months to 49.2 months has been found for the mean duration of imprisonment, mu, of political prisoners of a certain country with chronic PTSD.
A. The margin of error is 15.8 months. 42.2-17.6/2=15.8 B. One can be 95% confident that the maximum error made in estimating mu by x overbar is as found in part (a). C.The required sample size to have a margin of error of 12 months with a 99% confidence level is 82 prisoners. o= 42 months E=12 a=.01 Z.005= 2.576 (2.576*42/12)^2=82 D. The 99% confidence interval for the mean duration of imprisonment, mu, is from 24.5 months to 48.3 months. Z interval Stats o=42 Xbar= 36.4 n= 82 C-level= .99
A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects. Before treatment, 25 subjects had a mean wake time of 102.0 min. After treatment, the 25 subjects had a mean wake time of 75.7 min and a standard deviation of 21.7 min. Assume that the 25 sample values appear to be from a normally distributed population and construct a 99% confidence interval estimate of the mean wake time for a population with drug treatments. What does the result suggest about the mean wake time of 102.0 min before the treatment? Does the drug appear to be effective?
Construct the 99% confidence interval estimate of the mean wake time for a population with the treatment. 63.6 minless than mu less than 87.8 min (Round to one decimal place as needed.) What does the result suggest about the mean wake time of 102.0 min before the treatment? Does the drug appear to be effective? The confidence interval does not include the mean wake time of 102.0 min before the treatment, so the means before and after the treatment are different. This result suggests that the drug treatment has a significant effect.
Student t distribution def
The number of degrees of freedom for a collection of sample data is the number of sample values that can vary after certain restrictions have been imposed on all data values
Which of the following calculations is NOT derived from the confidence interval?
The population mean, mu= (Upper C-limit)+(Lower confidence limit)
student t distribution
Requirments: 1)Sample is a simple random sample 2) Either the sample is from a normally distrubted pop or more than 30 Sample mean is going to be the point measurement for the populaion mean If pop has a normal distribution of t= (xbar - mu)/(s/sqrt(n)) xbar= sample mean mu= population mean s= sample standard deviation n= dample size
The _______ is the best point estimate of the population mean.
Sample mean
In a test of the effectiveness of garlic for lowering cholesterol, 42 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes (beforeminusafter) in their levels of LDL cholesterol (in mg/dL) have a mean of 3.4 and a standard deviation of 19.3. Construct a 90% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment. What does the confidence interval suggest about the effectiveness of garlic in reducing LDL cholesterol?
T interval in calc n=42 xbar= 3.4 Sx= 19.3 C-level= .90 The confidence interval limits contain 0, suggesting that the garlic treatment did not affect the LDL cholesterol levels.
Which of the following is NOT a property of the Student t distribution?
The standard deviation of the Student t distribution is greater than 1, unlike the standard normal distribution, which has a standard deviation of 1.
A food safety guideline is that the mercury in fish should be below 1 part per million (ppm). Listed below are the amounts of mercury (ppm) found in tuna sushi sampled at different stores in a major city. Construct a 90% confidence interval estimate of the mean amount of mercury in the population. Does it appear that there is too much mercury in tuna sushi?
Tinterval L1 C-level .9 frequency 1 Yes, because it is possible that the mean is greater than 1 ppm. Also, at least one of the sample values exceeds 1 ppm, so at least some of the fish have too much mercury.
Which of the following would be a correct interpretation of a 99% confidence interval such as
We are 99% confident that the interval from 4.1 to 5.6 actually does contain the true value of mu. If we were to select many different samples of the same size and construct the corresponding confidence intervals, in the long run, 99% of them would actually contain the value of mu.
Refer to the technology output given to the right that results from measured hemoglobin levels (g/dL) in 100 randomly selected adult females. The results to the right are based on a 95% confidence level. Write a statement that correctly interprets the confidence level. TInterval (13.083, 13.549) x overbar=3.316 Sx=1.175 n=100
We have 95% confidence that the limits of 12.757 g/dL and 13.287 g/dL contain the true value of the mean hemoglobin level of the population of all adult females.
Refer to the technology output given to the right that results from measured hemoglobin levels (g/dL) in 100 randomly selected adult females. The confidence level of 99% was used. TInterval (13.178, 13.788) xoverbar=13.483 Sx=1.163 n=100
a. Express the confidence interval in the format that uses the "less than" symbol. Assume that the original listed data use two decimal places, and round the confidence interval limits accordingly. mu is greater than 13.178 and less than 13.788 b. Identify the best point estimate of mu and the margin of error. 13.483 The margin of error is .305 13.788-13.178/2 C. Since the sample size is greater than 30, the sample data do not need to be from a population with a normal distribution.
A data set includes 106 body temperatures of healthy adult humans having a mean of 98.9degreesF and a standard deviation of 0.63degreesF. Construct a 99% confidence interval estimate of the mean body temperature of all healthy humans. What does the sample suggest about the use of 98.6degreesF as the mean body temperature?
mu is greater than 98.739 and less than 99.061 This suggests that the mean body temperature could be higher than 98.6degreesF.
Margin of error estimate of mu (with o not known)
normal distribution E=t subscript alpha/2 (s/sqrt(n)) Where t a/2 has n-1 degrees of freedom. Table A-3 lists values of t a/2
An IQ test is designed so that the mean is 100 and the standard deviation is 20 for the population of normal adults. Find the sample size necessary to estimate the mean IQ score of nurses such that it can be said with 95% confidence that the sample mean is within 8 IQ points of the true mean. Assume that sigmaequals20 and determine the required sample size using technology. Then determine if this is a reasonable sample size for a real world calculation.
o= 20 E=8 Z=1.96 Sx=20 (1.96*20/8)^2= 24.01 round up to 25 Yes. This number of IQ test scores is a fairly small number.
Assume that we want to construct a confidence interval. Do one of the following, as appropriate: (a) find the critical value t Subscript alpha divided by 2, (b) find the critical value z Subscript alpha divided by 2, or (c) state that neither the normal distribution nor the t distribution applies. Here are summary statistics for randomly selected weights of newborn girls: n=244, x overbar=32.1 hg, s=7.8 hg. The confidence level is 95%
t=1.98
Confidence interval for the estimate of u
xbar - E greater than mu less than xbar + E
