Question

A survey of 100 randomly selected Michigan residents finds that the they spend a mean of $21 per month on going to the movies. An independent study of a 100 randomly selected California residents find that they spend a mean of $46 per month on going to the movies. Assume that the population standard deviation for the Michigan residents is $5 per month and that the population standard deviation for the California residents is $23 per month. Answer the following questions:

(a) If you were asked to construct a confidence interval what are the two pieces of inform you need?

(b) What is the point estimate for the above scenario?

(c) Calculate the Margin of Error at a 90% confidence level.

(d) Construct the Confidence Interval for the true difference in mean amount of money spent on going to the movies per month for the two states.

(e) Give interpretation of the Confidence Interval

Answer #1

Part a)

We need mean ( ) ± Margin of error

Part b)

Part c)

Margin of Error =
= 3.8719

Part d)

Confidence interval :-

Z(α/2) = Z (0.1 /2) = 1.645

Lower Limit =

Lower Limit = -28.8719

Upper Limit =

Upper Limit = -21.1281

**90% Confidence interval is ( -28.8719 , -21.1281
)**

Part e)

We are 90% confidence that the true population mean difference in amount of money spent on going to the movies per month for the two states. lies within the interval.

A survey of 100 randomly selected Michigan homeowners finds that
they spend a mean of $67 per month on home maintenance. In an
independent survey, 100 randomly selected Florida homeowners find
that they spend a mean of $75 per month on home maintenance.
Construct a 99% confidence interval for the true difference in the
mean amounts of money spent per month on home maintenance for the
two states. Assume that the population standard deviation is $14
per month for Michigan...

A survey of 87 randomly selected homeowners finds that they
spend a mean of $65 per month on home maintenance. Assume that the
population standard deviation is $13 per month.
Construct and interpret a 97% confidence interval for the mean
amount of money spent per month on home maintenance by all
homeowners.

A survey of 90 randomly selected homeowners finds that they
spend a mean of $66 per month on home maintenance. Assume that the
population standard deviation is $13 per month. Construct and
interpret a 95% confidence interval for the mean amount of money
spent per month on home maintenance by all homeowners.

A survey of 20 randomly selected adult men showed that the mean
time they spend per week watching sports on television is 9.34
hours with a standard deviation of 1.34 hours.
Assuming that the time spent per week watching sports on
television by all adult men is (approximately) normally
distributed, construct a 90 % confidence interval for the
population mean, μ .
Round your answers to two decimal places.
Lower bound: Enter your answer; confidence interval, lower
bound
Upper bound:...

The California Health Survey involved 51,048 randomly selected
adults in California. The author selected 100 of those subjects and
obtained these results. Among the 53 females, the mean height is
63.7 inches and the standard deviation is 4.5 inches. Among the 47
males, the mean is 69.2 inches and the standard deviation is 3.1
inches.
The author selected subjects from the California Health Survey
and 53 were female. Construct a 95% confidence interval for the
percentage of adults Californians that...

25 randomly selected students took a survey on how long they
spent on social media. The mean number of minutes these students
spent on social media was 205.1 minutes with a standard deviation
of 150 minutes.The population of the time spent on social media is
normally distributed. Construct a 98%confidence interval for the
population standard deviation of the time students spend on social
media.

A survey of 185 randomly selected homeowners found that
they spend a mean of $67 per month on home maintenance, with a
standard deviation of $14 per month. Estimate the mean amount of
money that a homeowner can expect to spend monthly on maintenance
at a 95% confidence level. You do not need to check conditions. Be
sure to interpret your answer in a sentence.

A mayor of a big city finds that 212 of 400 randomly
selected city residents believe that the city should spend more
money on road maintenance. Use this information for a large-sample
inference of the proportion of all city residents who believe that
the city should be spending more.
d) (4 points) Based on your findings, is it safe to say
that a majority of the city’s population wants more money spent on
roads? In other words, can you reject...

A random sample of 58 randomly selected people were asked how
much they spend on their grocery bill each week. The mean was
$115.36 with a standard deviation of $16.6. Find a 95% confidence
interval for the true population mean amount spent on groceries
each week. Report the answer accurate to four decimal places.

A sample of 105 laramie residents was randomly selected and
surveyed about Easter weekend spending. Their 95% confidence
interval for the true average amount of money that Laramie
residents spend during the Easter weekend was calculated to be
$277< μ < $327. National data shows that the
standard deviation of money spent duing easter weekend is
$91.00.
a.) What is the error?
b.) If I wanted to calculate a new confidence interval and drop
my error to $12.00, how large...

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 3 minutes ago

asked 23 minutes ago

asked 37 minutes ago

asked 37 minutes ago

asked 38 minutes ago

asked 47 minutes ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago