Question
In: Statistics and Probability
The average salary for a random sample of 49 new (less than 3 years) professors in...
The average salary for a random sample of 49 new (less than 3 years) professors in psychology graduate departments in 2012 was $92,000. The population, all professors with all years of experience in psychology graduate departments across the country, earned an average salary of $111,960, with a standard deviation of $40,295. Did the average salary of the new full psychology professors differ from that of all psychology professors as a whole? Conduct the appropriate statistical test to answer this question. Compute the 95% confidence interval for new professors.
a. (3 points) What is the confidence interval?
b. (2 points) How would you use this confidence interval to conduct the hypothesis test that you did for #3? That is, describe how you would reach the same conclusion by using the confidence interval
Solutions
orchestra answered 2 years agoRelated Solutions
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...
A random sample of 165 accountants and auditors had an average salary of $70,130 and a...
A random sample of 165 accountants and
auditors had an average salary of $70,130 and a standard deviation
in salary of $14,400. Another random sample of 182 loan officers
had an average salary of $67,960 and a standard deviation in salary
of $13,600. Construct a 98% confidence interval for the difference
in mean salary between the two professions.
Professors at a local university earn an average salary of $80,000 with a standard deviation of...
Professors at a local university earn an average salary of
$80,000 with a standard deviation of $6,000. The salary
distribution is approximately bell-shaped. What can be said about
the percentage of salaries that are less than $68,000 or more than
$92,000?
a. It is about 5%.
b. It is about 32%.
c. It is about 68%.
d. It is about 95%.
A random sample of 20 college professors is selected from all professors at a university. The...
A random sample of 20 college professors is selected from all
professors at a university. The following list gives their ages:
27, 52, 49, 63, 60, 57, 48, 53, 47, 42, 51, 54, 62, 60, 49, 58, 56,
47, 44, 49 [a] (1 point) What is the POPULATION in this problem?
[b] (1 point) What is the SAMPLE? [c] (1 point) State whether the
ages of professors are QUANTITATIVE or QUALITATIVE? [d] (1 point)
If quantitative, is it DISCRETE or...
A nationwide random sample of university professors revealed that 30 professors consumed a total of 7,500...
A nationwide random sample of university professors revealed
that 30 professors consumed a total of 7,500 mg of caffeine daily,
with a standard deviation of 40 mg. Assuming that the amount of
caffeine consumed per person per day is normally distributed, an
administrator wants to know if there is enough evidence to conclude
that the average amount of caffeine consumed per day by university
professors is more than 240 mg using α = 0.05?
Hypothesis test H0: Versus H1:
Using...
Is there enough evidence to conclude that the average salary at the Company A is LESS...
Is there enough evidence to conclude that the average salary at
the Company A is LESS than the average salary at the Company B ( α
= 0.04)?
Average Salary
Company A
Company B
$58,000
$47,000
$59,000
$74,000
$60,000
$56,000
$73,000
$55,000
$63,000
$65,000
$62,000
$52,000
$59,000
$68,000
$55,000
$63,000
$61,000
$66,000
$60,000
$66,000
$54,000
$57,000
$66,000
$59,000
$61,000
$58,000
$48,000
$63,000
$61,000
$66,000
$60,000
$62,000
$54,000
$62,000
$67,000
$53,000
$63,000
$46,000
$53,000
$54,000
$55,000
$65,000
$52,000
$61,000
$49,000
$72,000...
2.In a sample of 15 random people from a small town, the average salary is $25,778...
2.In a sample of 15 random people from a small town, the average
salary is $25,778 with a
standard deviation of $11,540. Use a 5% significance level to test
the claim that mean
salary in this town is less than $30,000
1. A random sample of 100 construction workers found that the average monthly salary was $21,310...
1. A random sample of 100 construction workers found
that the average monthly salary was $21,310 USD with a standard
deviation of $2405. A random sample of 81 Office workers found that
the average salary was $27,041 with a standard deviation of
$4651.
a.) Calculate a 95% confidence interval for the true
mean difference between office workers’ salaries and construction
workers’ salaries.
b.) Verify your conditions for inference
c.) Interpret your confidence interval with a sentence
in context.
d.) According...
A random sample of 23 chemists from Washington state shows an average salary of $46066 with...
A random sample of 23 chemists from Washington state shows an
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random sample of 21 chemists from Florida state shows an average
salary of $40860 with a standard deviation of $772. A chemist that
has worked in both states believes that chemists in Washington make
a different amount than chemists in Florida. At αα=0.05 is this
chemist correct? Let Washington be sample 1 and Florida be sample
2.
The...
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