Question
A sample of 53 night-school students' ages is obtained in order to estimate the mean age...
A sample of 53 night-school students' ages is obtained in order to estimate the mean age of night-school students. x = 25.6 years. The population variance is 27.
(a) Give a point estimate for μ. (Give your answer correct to one decimal place.)
(b) Find the 95% confidence interval for μ. (Give your answer correct to two decimal places.)
Lower Limit
Upper Limit
(c) Find the 99% confidence interval for μ. (Give your answer correct to two decimal places.)
Lower Limit
Upper Limit
Homework Answers
Answer #1
n=53, 
= 25.6 ,
=27
, 
=5.1962,
C=95%
a) Point Estimate = = 25.6
b) formula for Confidence Interval is
where Zc is the Z critical value for C= 95%
Zc= 1.96
24.201< 
<26.999
(24.20 , 27.00)
Lower Limit = 24.20
Upper Limit = 27.00
c)
formula for Confidence Interval is
where Zc is the Z critical value for C= 99%
Zc= 2.58
23.761< <27.439
Lower Limit = 23.76
Upper Limit = 27.44
Not the answer you're looking for?
Similar Questions
A college admissions director wishes to estimate the mean age of all students currently enrolled. In...
A college admissions director wishes to estimate the mean age of
all students currently enrolled. In a random sample of 19 students,
the mean age is found to be 23.7 years. From past studies, the ages
of enrolled students are normally distributed with a standard
deviation of 10.4 years. Construct a 90% confidence interval for
the mean age of all students currently enrolled.
1. The critical value:
2. The standard deviation of the sample mean:
3. The margin of error...
In a sample of 52 randomly selected students, 27 favored the amount being budgeted for next...
In a sample of 52 randomly selected students, 27 favored the
amount being budgeted for next year's intramural and
interscholastic sports. Construct a 99% confidence interval for the
proportion of all students who support the proposed budget amount.
(Give your answers correct to three decimal places.)
Lower Limit
Upper Limit
You may need to use the appropriate table in Appendix B to
answer this question.
Construct 90%, 95%, and 99% confidence intervals to estimate μ from the following data. State the...
Construct 90%, 95%, and 99% confidence intervals to estimate μ
from the following data. State the point estimate. Assume the data
come from a normally distributed population. 13.3 11.6 11.9 12.2
12.5 11.4 12.0 11.7 11.8 13.3
(Round the intermediate values to 4 decimal places. Round your
answers to 2 decimal places.)
90% confidence interval: enter the lower limit of the 90%
confidence interval ≤ μ ≤ enter the upper limit
of the 90% confidence interval
95% confidence interval: enter the...
Construct 90%, 95%, and 99% confidence intervals to estimate μ from the following data. State the...
Construct 90%, 95%, and 99% confidence intervals to estimate μ
from the following data. State the point estimate. Assume the data
come from a normally distributed population. 13.1 11.6 11.9 12.0
12.5 11.4 12.0 11.7 11.8 13.1 Appendix A Statistical Tables (Round
the intermediate values to 4 decimal places. Round your answers to
2 decimal places.) 90% confidence interval: enter the lower limit
of the 90% confidence interval ≤ μ ≤ enter the upper limit of the
90% confidence interval...
Construct 90%, 95%, and 99% confidence intervals to estimate μ from the following data. State the...
Construct 90%, 95%, and 99% confidence intervals to estimate μ
from the following data. State the point estimate. Assume the data
come from a normally distributed population. 12.0 11.6 11.9 12.9
12.5 11.4 12.0 11.7 11.8 12.0 Appendix A Statistical Tables (Round
the intermediate values to 4 decimal places. Round your answers to
2 decimal places.) 90% confidence interval: enter the lower limit
of the 90% confidence interval ≤ μ ≤ enter the upper limit of the
90% confidence interval...
Construct 90%, 95%, and 99% confidence intervals to estimate μ from the following data. State the...
Construct 90%, 95%, and 99% confidence intervals to estimate μ
from the following data. State the point estimate. Assume the data
come from a normally distributed population. 13.7 11.6 11.9 13.0
12.5 11.4 12.0 11.7 11.8 13.7 Appendix A Statistical Tables (Round
the intermediate values to 4 decimal places. Round your answers to
2 decimal places.) 90% confidence interval: enter the lower limit
of the 90% confidence interval ≤ μ ≤ enter the upper limit of the
90% confidence interval...
Ages of students: A simple random sample of 100 U.S. college students had a mean age...
Ages of students: A simple random sample of 100
U.S. college students had a mean age of 22.68 years. Assume the
population standard deviation is σ = 4.74 years. Construct
a 99% confidence interval for the mean age of U.S. college
students.
The answer I posted my instructor says it is wrong. I came up
with 21.45 to 23.91 a 99% confidence interval for the mean of the
U.S college students
She said the z procedures are needed.
Thanks.
1A) An experiment was designed to estimate the mean difference in weight gain for pigs fed...
1A) An experiment was designed to estimate the mean difference
in weight gain for pigs fed ration A as compared with those fed
ration B. Eight pairs of pigs were used. The pigs within each pair
were litter-mates. The rations were assigned at random to the two
animals within each pair. The gains (in pounds) after 45 days are
shown in the following table. Assuming weight gain is normal, find
the 95% confidence interval estimate for the mean of the...
question 1. A marketing specialist wants to estimate the average amount spent by visitors to an...
question 1. A marketing specialist wants to
estimate the average amount spent by visitors to an online
retailer's newly-designed website. From the data in a preliminary
study she guesses that the standard deviation of the amount spent
is about 16 dollars.
How large a sample should she take to estimate the mean amount
spent to within 4 dollars with 95% confidence? (Round your answer
up to the next largest integer).
______
question 2. Pepsi wants to use this technique
to...
The mean age for all Foothill College students for a recent Fall term was 33.2. The...
The mean age for all Foothill College students for a recent Fall
term was 33.2. The population standard deviation has been pretty
consistent at 15. Suppose that twenty-five Winter students were
randomly selected. The mean age for the sample was 30.4. We are
interested in the true mean age for Winter Foothill College
students. Let X = the age of a Winter Foothill College student.
A. Construct a 95% Confidence Interval for the true mean age of
Winter Foothill College...
ADVERTISEMENT
Need Online Homework Help?
Get Answers For Free
Most questions answered within 1 hours.
ADVERTISEMENT
Active Questions
asked 18 minutes ago
asked 21 minutes ago
asked 35 minutes ago
asked 36 minutes ago
asked 37 minutes ago
asked 37 minutes ago
asked 44 minutes ago
asked 59 minutes ago
asked 1 hour ago
asked 1 hour ago
asked 1 hour ago
asked 1 hour ago