In: Statistics and Probability

Suppose you have calculated the width of the 90% confidence interval for customer the average waiting time as ± 10 minutes, based on 10 replications of your experiment. However, your boss isn’t satisfied with your choice of a 90% confidence level and wants you to run additional experiments to give him the average waiting time with a 99% confidence interval of ± 10 minutes. Approximately how many additional replications will be required? (HINT: You may assume that the sample variance doesn’t change significantly when you incorporate the data from the additional replications.)

Let X be the waiting time in minutes of any given customer.

Let s be the sample variation. The estimated value of the population variance is

The standard error of the mean for a sample of size n=10 is

The level of significance for 90% confidence interval is

The critical value of z is

which is same as

Using standard normal tables we get

The margin of error obtained for 90% confidence interval with n=10 sample is ± 10 minutes

The level of significance for 99% confidence interval is

The critical value of z is

which is same as

Using standard normal tables we get

Assuming that the sample variance remains the same, the estimate of the population standard deviation remains at

the estimate of the standard error of the mean for a sample of size n is

we want the margin of error obtained for 99% confidence interval with a sample of size n to be ± 10 minutes

Rounding this up, we need a sample of size n=25 to get the average waiting time with a 99% confidence interval of ± 10 minutes.

ans: We need to run 15 additional experiments to give him the average waiting time with a 99% confidence interval of ± 10 minutes.

Suppose you constructed a 90% confidence interval for the true
average GPA of California community college students. If all of the
sample data were to be held constant, how would this confidence
interval compare to an 80% confidence interval based on the same
data?
A random sample of 100 students was taken to determine interest
in switching to a quarter system instead of semesters. Out of the
100 students, 73 stated that they were interested in switching to a
quarter...

If you wanted to calculate a 90% confidence interval for the
difference in average
number of friendship contacts between primary aged boys and girls
and we are
pretending that df=12, what t scores would you use? (assuming equal
variances again)
A. ☐+/- 1.356
B. ☐+/- 2.681
C. ☐+/- 1.782
D. ☐+/- 2.179
E. ☐+/- 3.055
9. Suppose you calculated your 90% interval as described above and
your lower
confidence limit was
–2.75 and your upper confidence limit was 3.20. What...

You are interested in finding a 90% confidence interval for the
average number of days of class that college students miss each
year. The data below show the number of missed days for 14 randomly
selected college students. Round answers to 3 decimal places where
possible. 1 8 10 9 1 3 1 8 1 0 2 1 5 8 a. To compute the confidence
interval use a distribution. b. With 90% confidence the population
mean number of days of...

A student was asked to find a 90% confidence interval for widget
width using data from a random sample of size n = 20. Which of the
following is a correct interpretation of the interval 11.8 < μ
< 21.7?
Check all that are correct.
The mean width of all widgets is between 11.8 and 21.7, 90% of
the time. We know this is true because the mean of our sample is
between 11.8 and 21.7.
With 90% confidence, the...

As the sample size INCREASES for computing a confidence
interval, the width of the confidence interval DECREASES.
12
When the population standard deviation sigma is assumed known, a
confidence interval can assume NORMALITY of the SAMPLE MEAN if the
sample size is greater than 30.
12
A SYMMETRIC histogram implies the plotted variable is NORMALLY
distributed.
12
The goal when using confidence intervals is to have WIDE
INTERVALS to be assured that the interval contains the population
parameter.
12
A...

You are interested in finding a 90% confidence interval for the
average commute that non-residential students have to their
college. The data below show the number of commute miles for 11
randomly selected non-residential college students. Round answers
to 3 decimal places where possible. 25 21 26 6 25 14 26 24 7 10 14
a. To compute the confidence interval use a distribution. b. With
90% confidence the population mean commute for non-residential
college students is between and miles....

Part 1. Suppose you are told that a 95% confidence interval for
the average price of a gallon of regular gasoline in your state is
from $2.99 to $3.99. Use the fact that the confidence interval for
the mean is in the form x − E to x +
E to compute the sample mean and the maximal margin of
error E. (Round your answers to two decimal places.)
x = $
E = $
Part 2. Anystate Auto Insurance...

A sample is selected to find a 90% confidence interval for the
average starting salary. Here are the sample statistics: n = 31, x
̄ = $43, 780, s = $1, 600.
a). Find the t− score used in the calculation of the confidence
interval.
b). Build a 90% confidence interval for the mean starting
salary.
c). Based on the result of part b), could we make a conclusion
that the mean staring salary is below $45, 000? Explain your...

If you had found a 90% confidence interval for µ, how would it
have differed from the 95% confidence interval?

Please calculate the 90% Confidence Interval for the population
slope in the following scenario. Suppose that you collected data to
determine the relationship between the amount of time a person
spends online as an independent variable and the amount of money a
person spends online as the dependent variable. Use the following
data for questions 6, 7, 8, 9, & 10. The regression equation is
= 24 + 10.1x, where x represents the number of hours a person
spends online...

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