A corporation that maintains a large fleet of company cars for the use of its sales staff is interested in the mean distance driven monthly per salesperson. The following list gives the monthly distances in miles driven by a random sample of 10 salespeople. 2155, 2164, 2081, 2253, 2520, 2490, 2584, 2463, 2278, 2085 Send data to calculator Send data to Excel Based on this sample, find a 95% confidence interval for the mean number of miles driven monthly by members of the sales staff, assuming that monthly driving distances are normally distributed. Give the lower limit and upper limit of the 95% confidence interval. Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. (If necessary, consult a list of formulas.) Lower limit: Upper limit: X 5

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Confidence interval for the population mean: Use of the t
A corporation that maintains a large fleet of company cars for the use of its sales staff is interested in the mean distance
driven monthly per salesperson. The following list gives the monthly distances in miles driven by a random sample of 10
salespeople.
2155, 2164, 2081, 2253, 2520, 2490, 2584, 2463, 2278, 2085
Send data to calculator
Send data to Excel
Based on this sample, find a 95% confidence interval for the mean number of miles driven monthly by members of the sales
staff, assuming that monthly driving distances are normally distributed. Give the lower limit and upper limit of the 95%
confidence interval.
Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. (If
necessary, consult a list of formulas.)
Lower limit:
Upper limit:
Transcribed Image Text:A corporation that maintains a large fleet of company cars for the use of its sales staff is interested in the mean distance driven monthly per salesperson. The following list gives the monthly distances in miles driven by a random sample of 10 salespeople. 2155, 2164, 2081, 2253, 2520, 2490, 2584, 2463, 2278, 2085 Send data to calculator Send data to Excel Based on this sample, find a 95% confidence interval for the mean number of miles driven monthly by members of the sales staff, assuming that monthly driving distances are normally distributed. Give the lower limit and upper limit of the 95% confidence interval. Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. (If necessary, consult a list of formulas.) Lower limit: Upper limit:
Done
=
O CONFIDENCE INTERVALS AND HYPOTHESIS TESTING
Confidence interval for the population mean: Use of the t...
Evelyn V
0/5
from
population we cannot assume to be approximately nommen. An Ciner Español
case, uppose we want to construct a confidence interval for the unknown
population mean μ, but we do not know the population standard deviation.
Then this confidence interval takes the following form.
point estimate(critical value standard error)
Here is another way to write the confidence interval.
point estimate-(critical value-standard error)<µ<point estimate+ (critical value-standard error)
Since we do not know the value of o, we use the sample standard deviation
s to approximate it. Then for our confidence interval, the point estimate is
the sample mean and the standard error is where is the sample size.
For the critical value, we use where the t distribution has "-1 degrees of
Ž
freedom and a-(1-confidence level). This information is summarized below.
point estimate: degrees of freedom: n-1 critical value: standard error:
x
www-awu.aleks.com
So, we can write the confidence interval in this form. (When we use the
formula we remember that the distribution has n-1 degrees of freedom.)
¯¯(';¯†)<*~*• (';¨¯†)
x=4.752
x=4.752
The current problem:
For our problem we can assume that the population is (approximately)
normal and we do not know the population standard deviation G. We want to
find the 95% confidence interval for the population mean . Here the sample
mean, the sample standard deviations, and the sample size n are as
follows.
n=17
So, here are the point estimate, the number of degrees of freedom, the
critical value, and the standard error.
Point estimate Degrees of freedom
s 0.057
4.752-
52- (2.120.0057)<μ<
Lower limit: 4.72
Upper limit: 4.78
n-1-17-1-16
More Practice
Deriving this formula
<μ<4.752-
Finding X and S
AA
More
Critical value
(1-0.95) -0.025 2.120
Using those values, here is how we calculate the confidence interval.
2+ (2.120 -0.057)
Rounding to two decimal places, here is what we get.
Standard error
0.057
√ √17
4.72<µ<4.78
The value 4.72 is the approximate lower limit and the value 4.78 is the
approximate upper limit.
ANSWER
M
P
B
6
© 2023 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center
Transcribed Image Text:Done = O CONFIDENCE INTERVALS AND HYPOTHESIS TESTING Confidence interval for the population mean: Use of the t... Evelyn V 0/5 from population we cannot assume to be approximately nommen. An Ciner Español case, uppose we want to construct a confidence interval for the unknown population mean μ, but we do not know the population standard deviation. Then this confidence interval takes the following form. point estimate(critical value standard error) Here is another way to write the confidence interval. point estimate-(critical value-standard error)<µ<point estimate+ (critical value-standard error) Since we do not know the value of o, we use the sample standard deviation s to approximate it. Then for our confidence interval, the point estimate is the sample mean and the standard error is where is the sample size. For the critical value, we use where the t distribution has "-1 degrees of Ž freedom and a-(1-confidence level). This information is summarized below. point estimate: degrees of freedom: n-1 critical value: standard error: x www-awu.aleks.com So, we can write the confidence interval in this form. (When we use the formula we remember that the distribution has n-1 degrees of freedom.) ¯¯(';¯†)<*~*• (';¨¯†) x=4.752 x=4.752 The current problem: For our problem we can assume that the population is (approximately) normal and we do not know the population standard deviation G. We want to find the 95% confidence interval for the population mean . Here the sample mean, the sample standard deviations, and the sample size n are as follows. n=17 So, here are the point estimate, the number of degrees of freedom, the critical value, and the standard error. Point estimate Degrees of freedom s 0.057 4.752- 52- (2.120.0057)<μ< Lower limit: 4.72 Upper limit: 4.78 n-1-17-1-16 More Practice Deriving this formula <μ<4.752- Finding X and S AA More Critical value (1-0.95) -0.025 2.120 Using those values, here is how we calculate the confidence interval. 2+ (2.120 -0.057) Rounding to two decimal places, here is what we get. Standard error 0.057 √ √17 4.72<µ<4.78 The value 4.72 is the approximate lower limit and the value 4.78 is the approximate upper limit. ANSWER M P B 6 © 2023 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center
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