Statistics Homework 12: Ch. 7 & 8
take the max-min then divide by 2. EX: 47.3-19.1/2
A 95% confidence interval of 19.1 months to 47.3 months has been found for the mean duration of imprisonment, μ, of political prisoners of a certain country with chronic PTSD. a. Determine the margin of error, E.
One can be 95% confident that the maximum error made in estimating μ by x is as found in part (a).
A 95% confidence interval of 19.1 months to 47.3 months has been found for the mean duration of imprisonment, μ, of political prisoners of a certain country with chronic PTSD. b. Explain the meaning of E in this context in terms of the accuracy of the estimate.
First find the t score. use 1-99% to get 0.01 and N is 110-1 T.INV.2T(0.01, 109) Then take that score and times it by the SD. then divide it by sqrt of n or 110. using that answer take the mean and add it to that. that is the max. then minus it from the mean. that is the min. example: =98-0.185 & =98+0.185
A data set includes 110 body temperatures of healthy adult humans having a mean of 98.0°F and a standard deviation of 0.74°F. Construct a 99% confidence interval estimate of the mean body temperature of all healthy humans. What does the sample suggest about the use of 98.6°F as the mean body temperature?
To find sample size you need to first find Z score. Since we have probability divide that by 2. Ex: 0.05/2= 0.025. and its to the right use =ABS(NORM.S.INV(0.025). then =roundup((Z score*SD/error0^2, rounded up) 0. Ex: =Roundup((1.956*10/2)^2,0)
An IQ test is designed so that the mean is 100 and the standard deviation is 10 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 2 IQ points of the true mean. Assume that σ=10 and determine the required sample size using technology. Then determine if this is a reasonable sample size for a real world calculation.
Yes. This number of IQ test scores is a fairly small number.
An IQ test is done... Then determine if this is a reasonable sample size for a real world calculation..
take the amount of months and add or minus the Z score that you got doing =norm.s.inv example: =36.2-2.575*42/sqrt(70) where 42 is SD and 70 is N then do the same but adding to get max.
Find a 99% confidence interval for the mean duration of imprisonment, μ, if a sample of the size determined in part (c) has a mean of 36.2 months.
Find probability by taking percentage and minus 1. Example: 95%-1= 0.05 then use =T.INV.2T(0.05,207) 208-1=207 that is N.
Here are summary statistics for randomly selected weights of newborn girls: n=208, x=33.4 hg, s=7.4 hg. The confidence level is 95%.
Since the sample size is greater than 30, the sample data do not need to be from a population with a normal distribution.
In constructing the confidence interval estimate of μ, why is it not necessary to confirm that the sample data appear to be from a population with a normal distribution? Choose the correct answer below.
The sample size must be greater than 30
In order to use the Normal Distribution:
We have 90% confidence that the limits of 12.851 g/dL and 13.245 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 results to the right are based on a 90% confidence level. Write a statement that correctly interprets the confidence level.
Use the term X that is Given
The best point estimate of μ is
Sample mean
The _______ is the best point estimate of the population mean.
The population mean, μ= min+max
Which of the following calculations is NOT derived from the confidence interval?
The standard deviation of the Student t distribution is s=1.
Which of the following is NOT a property of the Student t distribution?
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.
Which of the following is NOT a requirement for constructing a confidence interval for estimating a population mean with σ known?
161.7±27.8
Which of the following is NOT an equivalent expression for the confidence interval given by 161.7<μ<189.5?
We are 99% confident that the interval from 4.1 to 5.6 actually does contain the true value of μ.
Which of the following would be a correct interpretation of a 99% confidence interval such as 4.1<μ<5.6?
=(1-0.99)/2 Norm.S.INV(1-0.005) =Roundup((Above answer*SD/error)^2,0)
c. Find the sample size required to have a margin of error of 11 months and a 99% confidence level. (Use σ=38 months.)
First find the mean/AVERAGE and =STDEV.S for the sample. then find T score using =T.INV.2T(x, n) x being percentage minus one and n being sample size minus one. then * by SD and then divide by sqrt of n. then add and minus to mean. that gives you max/min
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 95% confidence interval estimate of the mean amount of mercury in the population. Does it appear that there is too much mercury in tuna sushi?
max-min the divide by two
margin of error is
