Question
In: Statistics and Probability
A random sample of n1 = 261 people who live in a city were selected and...
A random sample of n1 = 261 people who live in a city were
selected and 77 identified as a "dog person." A random sample of n2
= 101 people who live in a rural area were selected and 52
identified as a "dog person." Find the 95% confidence interval for
the difference in the proportion of people that live in a city who
identify as a "dog person" and the proportion of people that live
in a rural area who identify as a "dog person."
Round answers to to 4 decimal places.
? < p1 - p2 < ?
Solutions
orchestra answered 1 year agoRelated Solutions
a random sample of n1= 156 people ages 16 to 19 were taken from the island...
a
random sample of n1= 156 people ages 16 to 19 were taken from the
island of oahu, hawaii, and 16 were found to be high school
dropouts. Another random sample of n2=129 people ages 16 to 19 werr
taken from sweetwater county, Wyoming, and 10 were found to be high
school dropouts. Do these data indicate that the population
proportion of high school dropouts on Oahu is different ( either
way) from that of sweetwater county? use a 1%...
A sample of 1000 people that live in a large city was asked if they believed...
A sample of 1000 people that live in a large city was asked if
they believed that the city’s sales tax proposal was warranted. If
at least two-thirds of the city’s citizens believe that the sales
tax is warranted, then the mayor will take immediate steps to
formally propose the tax. The survey responses showed that 692 of
the 1000 people surveyed believe that the tax proposal is
warranted.
Write the appropriate null and alternative hypotheses for the
one-tailed test...
A sample of 1000 people that live in a large city was asked if they believed...
A sample of 1000 people that live in a large city was asked if
they believed that the city’s sales tax proposal was warranted. If
at least two-thirds of the city’s citizens believe that the sales
tax is warranted, then the mayor will take immediate steps to
formally propose the tax. The survey responses showed that 692 of
the 1000 people surveyed believe that the tax proposal is
warranted.
Find the critical value for α=0.1.
What is the result of...
Suppose a random sample of 100 households were selected from a large city. if the true...
Suppose a random sample of 100 households were selected from a
large city. if the true proportion of households in the city that
owns at least one pet is 0.4, then what is the probability that at
least 50% of the sample owns at least one pet?
a) 0.0207
b) 0.9793
c) 0.5793
d) 0.4207
e) 0.2578
A random sample of size n1 = 16 is selected from a normal population with a...
A random sample of size n1 = 16 is selected
from a normal population with a mean of 74 and a standard deviation
of 9. A second random sample of size n2 = 7 is
taken from another normal population with mean 68 and standard
deviation 11. Let X1and X2 be the two sample means. Find:
(a) The probability that X1-X2 exceeds 4.
(b) The probability that 4.8 ≤X1-X2≤ 5.6.
Round your answers to two decimal places (e.g. 98.76).
A random sample of size n1 = 14 is selected from a normal population with a...
A random sample of size n1 = 14 is selected
from a normal population with a mean of 74 and a standard deviation
of 6. A second random sample of size n2 = 9 is
taken from another normal population with mean 70 and standard
deviation 14. Let X¯1and X¯2 be the two sample means. Find:
(a) The probability that X¯1-X¯2 exceeds 3.
(b) The probability that 4.4 ≤X¯1-X¯2≤ 5.4.
In one of the studies, it was found that, in a random sample of 261 married...
In one of the studies, it was found that, in a random sample of
261 married persons, 135 were smokers while in a sample of 239
non-married persons there were 131 smokers.
a. Find a 90% confidence interval for the true difference in
proportion of smokers among the married and non-married
populations.
b. Based on the above interval, can one conclude that there is a
significant difference between the proportions of smokers in the
two populations? Justify your answer
c....
The mean age of a random sample of 25 people who were playing the slot machines...
The mean age of a random sample of 25 people who were playing
the slot machines is 55.3 years and the standard deviation is 6.8
years. The mean age of a random sample of 35 people who were
playing roulette is 49.7 years with a standard deviation of 4.2
years.
Find the 99% confidence interval of the true difference in
means.
Would you reject the null hypothesis that there is no
difference in the mean ages of slot machine players...
Dr. Morozov is interested in whether people who live in noisy parts of a city have...
Dr. Morozov is interested in whether people who live in noisy
parts of a city have worse hearing. To test this, she administers
hearing tests to 6 randomly selected healthy eighteen-year-olds who
have grown up in one of the noisiest parts of the city. On the
hearing test, they obtain a score of M = 17, s = 3.10. This test
was designed so that healthy 18-year-olds should score 20 (i.e.μ =
20, σ is unknown). Go through the hypothesis...
Independent random samples of n1 = 700 and n2 = 520 observations were selected from binomial...
Independent random samples of
n1 = 700
and
n2 = 520
observations were selected from binomial populations 1 and 2,
and
x1 = 335
and
x2 = 378
successes were observed.
(a) Find a 90% confidence interval for the difference
(p1 − p2) in the two
population proportions. (Round your answers to three decimal
places.)
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