Question
In: Statistics and Probability
The manager of a grocery store has taken a random sample of 100 customers. The average...
The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took the customers in the sample to check out was 4.8 minutes with a sample standard deviation of 0.5 minutes. Conduct a test to determine whether or not the mean waiting time of all customers is significantly more than 5 minutes. Use a significance level of α = 0.01.
Solutions
denotes the mean
waiting time of all customers.
orchestra answered 1 year agoRelated Solutions
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1. At store A, the sample average amount spent by a random
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significance, whether or not there is a difference between the
population mean amount spent per customer at store A and the...
A random sample of 29 lunch customers was taken at a restaurant The average amount of...
A random sample of 29 lunch customers was taken at a restaurant
The average amount of time the customers in the sample stayed in
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minutes.
a) Compute the standard error of the mean?
b) Construct a 68% confidence interval for the true average
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c) Construct a 90% confidence interval for the true average
amount of time customers spent in the restaurant?...
A random sample of 49 lunch customers was taken at a restaurant. The average amount of...
A random sample of 49 lunch customers was taken at a restaurant.
The average amount of time these customers stayed in the restaurant
was 45 minutes with a standard deviation of 14 minutes.
a. Compute the standard error of the mean.
b. Construct a 95% confidence interval for the true average
amount of time customers spent in the restaurant.
c. With a .95 probability, how large of a sample would have to
be taken to provide a margin of error...
A random sample of 49 lunch customers was taken at a restaurant. The average amount of...
A random sample of 49 lunch customers was taken at a
restaurant. The average amount of time these customers stayed in
the restaurant was 45 minutes with a sample standard deviation of
14 minutes.
Compute the standard error of the mean.
At 95% confidence, what is the confidence interval
estimate?
Construct a 99% confidence interval for the true average amount
of time customers spent in the restaurant.
Compare the margin of errors and width of your confidence
intervals from parts...
(Operations Management) The grocery store gets an average of 50 customers per shift. The manager wants...
(Operations Management) The grocery
store gets an average of 50 customers per shift. The manager wants
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salespeople. The average waiting times will be seven minutes with
one salesperson, four minutes with two salespeople, three minutes
with three salespeople, and two minutes with four salespeople. He
has estimated the cost per minute that customers wait at $1. The
cost per salesperson per shift (including fringe benefits) is $70.
Analyze to...
A random sample of 100 daily stock returns was taken. The average return in the sample...
A random sample of 100 daily stock returns was taken. The
average return in the sample appeared to be 4.81%. The population
standard deviation is not known, but the sample standard deviation
is known to be 0.9%. (show your work) a. Calculate a 95% confidence
interval for the average stock return b. How big the sample size be
for the margin of error of 0.1% at the confidence level 95%? c. A
fund manager claims that the average stock return...
A random sample of 100 daily stock returns was taken. The average return in the sample...
A random sample of 100 daily stock returns was taken. The
average return in the sample appeared to be 4.81%. The population
standard deviation is not known, but the sample standard deviation
is known to be 0.9%.
a. Calculate a 95% confidence interval for the
average stock return
b. How big the sample size be for the margin of error of 0.1% at
the confidence level 95%?
c. A fund manager claims that the average stock return is higher
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A random sample of 200 customers who visited a local grocery spent an average of $97....
A random sample of 200 customers who visited a local grocery
spent an average of $97. You find a reputable source stating that
the current population standard deviation of all such grocery bills
in similar grocery stores is equal to $19. Using this information,
find the 90% confidence interval for the population mean:
$95.28 to $98.72
$94.79 to $99.21
$94.37 to $99.63
$93.87 to $100.13
A random sample of 200 customers who visited a local grocery spent an average of $97....
A random sample of 200 customers who visited a local grocery
spent an average of $97. You find a reputable source stating that
the current population standard deviation of all such grocery bills
in similar grocery stores is equal to $19. Using this information,
find the 90% confidence interval for the population mean:
$95.28 to $98.72
$94.79 to $99.21
$94.37 to $99.63
$93.87 to $100.13
simple random sample of 80 customers is taken from a customer information file and the average...
simple random sample of 80 customers is taken from a customer
information file and the average age is 34. The population standard
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also calculated from the sample and is found to be 5.5. (6
points)
Test the hypothesis that the population mean age is greater
than 30 using the critical value approach and a 0.05 level of
significance.
Test the hypothesis that the population mean age is less than...
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