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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. 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: (642.4,64+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.
- 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. 1.282 1.645 1.960 2.326 2.
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
- 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).
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
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. 1.282 1.645 1.960 2.326 2.
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. 1.282 1.645 1.960 2.326 2.
Round the final answer to two decimal places.
(CL) is 90%, or 0.9. So, we can calculate alpha (±). ±=1C=L=1 =0. =0. Since ±=0.1,we know that ±2 =0. =0. 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. 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. 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. inches.
they have a pet, and 67 reported that they have a pet.
What value of z should be used to cal- culate a confidence interval with a 90% confidence level?
z0.10 z0.05 z0.025 z0.01 z0. 1.282 1.645 1.960 2.326 2.
10. The lengths, in inches, of adult corn snakes are normally distributed with a population standard deviation of 8 inches and an unknown population mean. A random sample of 25 snakes is taken and results in a sample mean of 58 inches. Identify the parameters needed to cal- culate a confidence interval at the 99% confidence level. Then find the confi- dence interval.
z0.10 z0.05 z0.025 z0.01 z0. 1.282 1.645 1.960 2.326 2.
11. Adult entrance fees to amusement parks in the United States are normal- ly distributed with a population stan- dard deviation of 2.5 dollars and an unknown population mean. A random
In this formula, z = z ±/2 =z0.025 =1. because the confidence level is 95%. Ã=5and ME =2. Therefore, n= z2Ã2M/ E =(1.96)^2(5)^2/2^ H24. Use n=25 to ensure that the sample size is large enough.
x = 58 Ã = 8 n = 25 z ±/2 = 2. (53.88, 62.12)
the given values Ã=8n, =25, and z ±/2=2.576for a confidence level of 99%, we have margin of error=(2.576)(8/ 25) H (2.576)(1.6) H4. With x¯=58 and a margin of error of 4.12, the confidence interval is (58 4.12,58+4.12) = (53.88 , 62.12).
We can estimate with 99% con- fidence that the true population mean length of adult corn snakes is between 53.88 and 62.12 inches.
The margin of error is given by margin of error = (z ±/2)(nÃ)/
sample of 22 entrance fees at different Substituting the given values Ã=2.5,
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. 1.282 1.645 1.960 2.326 2.
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. 1.282 1.645 1.960 2.326 2.
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.
Ã=
n= z ±/2 =1. confidence level of 90%, we find the margin of error = (1.645)(17/ 15) H (1.645)(4.389) H7. 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.
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.
z0.10 z0.05 z0.025 z0.01 z0. 1.282 1.645 1.960 2.326 2.
- 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.
