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.

You are interested in finding a 90% confidence interval for the
mean number of visits for physical therapy patients. The data below
show the number of visits for 13 randomly selected physical therapy
patients. Round answers to 3 decimal places where possible. 19 19
14 13 26 19 5 14 17 27 13 20 28 a. To compute the confidence
interval use a distribution. b. With 90% confidence the population
mean number of visits per physical therapy patient is between...

You are interested in finding a 90% confidence interval for the
mean number of visits for physical therapy patients. The data below
show the number of visits for 15 randomly selected physical therapy
patients. Round answers to 3 decimal places where possible.
13
26
16
13
6
27
26
6
13
25
6
11
20
14
16
a. To compute the confidence interval use a ? z
t distribution.
b. With 90% confidence the population mean number of visits per
physical...

You are interested in finding a 90% confidence interval for the
mean number of visits for physical therapy patients. The data below
show the number of visits for 12 randomly selected physical therapy
patients. Round answers to 3 decimal places where possible.
5
12
18
17
28
9
10
6
16
28
5
13
b. With 90% confidence the population mean number of visits per
physical therapy patient is between and
visits.
c. If many groups of 12 randomly selected physical...

10. Construct the 90% confidence interval for each of the
following scenarios. a. You wish to estimate the proportion of
automobiles that are purchased over the Internet. You took a random
sample of 200 people and found that 38 purchased a car over the
Internet. b. In a random sample of 17 bikers, the mean time spent
riding a day was 2.813 hours and the standard deviation was 1.205.
Assume the hours spent are normally distributed.

As a consultant, you need to use the Hospital database and
construct a 90% confidence interval to estimate the average census
for hospitals. Change the level of confidence to 99%. What happened
to the interval? Did the point estimate change?
Determine the sample proportion of the Hospital database under
the variable “service” that are “general medical” (category 1).
From this statistic, construct a 95% confidence interval to
estimate the population proportion of hospitals that are “general
medical.” What is the...

Answer in terms of Econometrics
1. A confidence interval for ?? is calculated to be (2, 6). Is
the parameter in the interval? Explain.
2. If the results of many regressions are to be reported, how do
you do it?

A)In our sample of 204 students, 27 have September birthdays.
Calculate a 90% confidence interval for the proportion of 332
students that have a September birthday.
B) Interpret the interval from part b.
C) If the birth months are equally likely (ignoring the
difference in the number of days), what proportion of births would
be in September?
D) For various reasons, September is considered to be the most
common birth month. Using your confidence interval from part A and
response...

You are interested in finding a 98% 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 10 randomly
selected college students.
5
4
10
8
1
2
6
5
8
11
a. To compute the confidence interval use a
distribution.
b. With 98% confidence the population meannumber of days of
class that college students miss is between and days.
c. If many groups...

A newsgroup is interested in constructing a 90% confidence
interval for the difference in the proportions of Texans and New
Yorkers who favor a new Green initiative. Of the 582 randomly
selected Texans surveyed, 428 were in favor of the initiative and
of the 569 randomly selected New Yorkers surveyed, 456 were in
favor of the initiative.
. With 90% confidence the difference in the proportions of
Texans and New Yorkers who favor a new Green initiative is between
(round...

SCENARIO 8-12The Three Brothers Energy Drink Company bottles and distributes a
popular drink for athletes and exercise enthusiasts. Because of its
marketing successes the company has installed an additional filling
machine and the managers are eager to use it in daily operations.
The machine is set to fill bottles at 16 oz.However, we know there is inherent machine variability and quality
control has determined through testing a mean of 16.2 oz. and a
standard deviation of 0.3 oz. using a...

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