Question

Suppose that we wish to assess whether more than 60 percent of all U.S. households in a particular income class bought life insurance last year. That is, we wish to assess whether p, the proportion of all U.S. households in the income class that bought life insurance last year, exceeds .60. Assume that an insurance survey is based on 1,000 randomly selected U.S. households in the income class and that 640 of these households bought life insurance last year. a) Assuming that p equals .60 and the sample size is 1,000, what is the probability of observing a sample proportion that is at least .64? b) Based on your answer in part a, do you think more than 60 percent of all U.S. households in the income class bought life insurance last year? Explain.

Answer #1

Suppose that we wish to assess whether more than 60 percent of
all U.S. households in a particular income class bought life
insurance last year. That is, we wish to assess whether p, the
proportion of all U.S. households in the income class that bought
life insurance last year, exceeds .60. Assume that an insurance
survey is based on 1,000 randomly selected U.S. households in the
income class and that 640 of these households bought life insurance
last year. 1....

Respond to the following in a minimum of 175 words:
Suppose that we wish to assess whether more than 60 percent of
all U.S. households in a particular income class bought life
insurance last year. That is, we wish to assess whether p, the
proportion of all U.S. households in the income class that bought
life insurance last year, exceeds .60. Assume that an insurance
survey is based on 1,000 randomly selected U.S. households in the
income class and that...

Suppose a national news agency reported that more than 88% of
all households purchased at least one item online last year. We
wish to test this statement and randomly sample 200 households.
When asked if they made any online purchases last year 180 reported
“YES”. At the 0.05 level of significance, does the sample data
support the statement made by the national news agency?
1. State the null and alternative hypotheses of interest. Ho
___________________________ HA ___________________________
2. What is...

uppose a national news agency reported that more than 88% of all
households purchased at least one item online last year. We wish to
test this statement and randomly sample 200 households. When asked
if they made any online purchases last year 180 reported “YES”. At
the 0.05 level of significance, does the sample data support the
statement made by the national news agency? 1. State the null and
alternative hypotheses of interest. Ho ___________________________
HA ___________________________
2. What is...

Research shows that even for married couples in two-income
households, women perform significantly more housework than men
(see Hersh and Stratton 2000). You wish to replicate this study in
2019 and you sample populations of married men and women, all of
whom work fulltime outside the home. Housework (Y) is measured in
hours and the standard deviations are significantly different from
one another. Do your data below support the previous research
findings (that women perform more housework)
Women
Men
1...

Thirty-eight percent of all Americans drink bottled water more
than once a week (Natural resources Defense Council, December 4,
2015). Suppose you have been hired by the Natural Resources Defence
Council to investigate bottled water consumption in St. Paul. You
plan to select a sample of St. Paulites to estimate the proportion
who drink bottled water more than once a week. Assume the
population proportion of St. Paulites who drink bottled water more
than once a week is , the...

1. We wish to estimate what percent of adult residents in a
certain county are parents. Out of 400 adult residents sampled, 296
had kids. Based on this, construct a 95% confidence interval for
the proportion p of adult residents who are parents in this county.
Express your answer in tri-inequality form. Give your answers as
decimals, to three places.
__< p <__ Express the same answer using the point estimate
and margin of error. Give your answers as decimals,...

Absentee rates - Friday vs Wednesday: We want
to test whether or not more students are absent on Friday afternoon
classes than on Wednesday afternoon classes. In a random sample of
302 students with Friday afternoon classes, 48 missed the class. In
a different random sample of 307 students with Wednesday afternoon
classes, 32 missed the class. The table below summarizes this
information. The standard error (SE) is given to save calculation
time if you are not using software.
Data...

Absentee rates - Friday vs Wednesday: We want
to test whether or not more students are absent on Friday afternoon
classes than on Wednesday afternoon classes. In a random sample of
302 students with Friday afternoon classes, 48 missed the class. In
a different random sample of 307 students with Wednesday afternoon
classes, 30 missed the class. The table below summarizes this
information. The standard error (SE) is given to save calculation
time if you are not using software.
Data...

1. A poll of 2234 U.S. adults found that 31% regularly used
Facebook as a news source.
Find the margin of error and confidence interval for the
percentage of U.S. adults who regularly use Facebook as a news
source, at the 90% level of confidence. Round
all answers to 2 decimal places.
Margin of Error (as a percentage):
Confidence Interval: % to %
Find the margin of error and confidence interval for the
percentage of U.S. adults who regularly use...

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