Question

In: Statistics and Probability

In a recent survey of drinking laws, a random sample of 1000 women showed that 65%...

In a recent survey of drinking laws, a random sample of 1000 women showed that 65% were in favor of increasing the legal drinking age. In a random sample of 1000 men , 60% favored increasing the legal drinking age. Test the hypothesis that the percentage of women favoring a higher legal drinking age is greater than the percentage of men. Use a= 0.05

Call women population 1 and men population 2

A) What initial conclusion would be reached in this hypothesis test?

a. Reject H1

b. Reject H0

c. Do not reject H0

d. Do not reject H1

B) What is the possible error type and the correct statement of the possible error?

a. Type 1: The sample data indicated that the proportion of women in favor of increasing the drinking age is greater than the proportion of men, but actually the proportion is less than or equal to.

b. Type 2: The sample data indicated that the proportion of women favoring a higher drinking age is equal to the proportion of men, but actually the proportion of women is greater.

c. Type 2: The sample data indicated that the proportion of women who favor a higher drinking age is less than the proportion of men, but actually the proportions are equal.

d. Type 1: The sample indicated that the proportion of women who favor a higher drinking age is greater than the proportion of men, but actually the proportion of men favoring a higher drinking age is greater.

C) State the conclusion in a sentence.

Solutions

Expert Solution

Given that,
sample one, x1 =650, n1 =1000, p1= x1/n1=0.65
sample two, x2 =600, n2 =1000, p2= x2/n2=0.6
null, Ho: p1 = p2
alternate, H1: p1 > p2
level of significance, α = 0.05
from standard normal table,right tailed z α/2 =1.645
since our test is right-tailed
reject Ho, if zo > 1.645
we use test statistic (z) = (p1-p2)/√(p^q^(1/n1+1/n2))
zo =(0.65-0.6)/sqrt((0.625*0.375(1/1000+1/1000))
zo =2.309
| zo | =2.309
critical value
the value of |z α| at los 0.05% is 1.645
we got |zo| =2.309 & | z α | =1.645
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value: right tail - Ha : ( p > 2.3094 ) = 0.01046
hence value of p0.05 > 0.01046,here we reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 > p2
test statistic: 2.309
critical value: 1.645
A.
decision: reject Ho
option:b
p-value: 0.01046
B.
a. Type 1: The sample data indicated that the proportion of women in favor of increasing the drinking age is greater than the proportion of men, but actually the proportion is less than or equal to.
C.
we have enough evidence to support the claim that the percentage of women favoring a higher legal drinking age is greater than the percentage of men


Related Solutions

3. In a recent survey of gun control laws, a random sample of 1000 women showed...
3. In a recent survey of gun control laws, a random sample of 1000 women showed that 65% were in favor of stricter gun control laws.  In a random sample of 1000 men, 60% favored stricter gun control laws.  Test the claim that the percentage of men and women favoring stricter gun control laws is the same at α=0.05 1 H0:_________   Ha:____________                       5.Decision:{Circle one}Reject H0 or Fail to Reject H0 2 α=_________                                                   6. P-value ____________ 3 Critical Value   _________                                 7.Statement:___________________________________                                                                                           ___________________________________ 4Test stat___________                                                                 
In a random sample of 360 women, 65% favored stricter gun control laws. In a random...
In a random sample of 360 women, 65% favored stricter gun control laws. In a random sample of 220 men, 60% favored stricter gun control laws. Test the claim that the proportion of women favoring stricter gun control is higher than the proportion of men favoring stricter gun control. Use a significance level of 0.05.
TWO PROPORTIONS In a random sample of 360 women, 65% favored stricter gun control laws. In...
TWO PROPORTIONS In a random sample of 360 women, 65% favored stricter gun control laws. In a random sample of 220 men, 60% favored stricter gun control laws. Test the claim that the proportion of women favoring stricter gun control is higher than the proportion of men favoring stricter gun control. Hypothesis Ho: Ha:                                                                  p-value                                                                            Conclusion .                                                                                             
A random survey of 1000 students nationwide showed a mean ACT score of 21.4, with a...
A random survey of 1000 students nationwide showed a mean ACT score of 21.4, with a standard deviation of 4. A survey of 500 randomly selected Ohio scores showed a mean of 20.8, with a population standard deviation of 3. At a=0.05, can we conclude that Ohio is below the national average?
Questions A - F A recent survey of 1000 American women between the ages of 45...
Questions A - F A recent survey of 1000 American women between the ages of 45 and 64 asked them what medical condition they most feared. Of those sampled 61% said breast cancer, 8% said heart disease, and the rest picked other conditions. By contrast, currently about 3% of female deaths are due to breast cancer, whereas 32% are due to heart disease. A. Construct a 95% confidence interval for the population proportion of women who most feared breast cancer....
A recent survey of 1000 adults between 18 and 40 showed that 34% said they had...
A recent survey of 1000 adults between 18 and 40 showed that 34% said they had no credit cards. Find the 99% confidence interval for the population proportion.
A recent survey showed that 65% of U.S. employers were likely to require higher employee contributions...
A recent survey showed that 65% of U.S. employers were likely to require higher employee contributions for health care coverage. Suppose the survey was based on a sample of 600 companies. Find the 98% confidence interval for the proportion of all companies likely to require higher employee contributions for health care coverage. Find left endpoint and right endpoint.
A random sample of 100 Ohio businesses showed that 34 were owned by women. A sample...
A random sample of 100 Ohio businesses showed that 34 were owned by women. A sample of 200 New Jersey businesses showed that 64 were owned by women. Construct a 90% confidence interval for the difference in proportions of women-owned business in the two states. Use u(Ohio - u(New Jersey)
In an opinion survey, a random sample of 1000 adults from the U.S.A. and 1000 adults...
In an opinion survey, a random sample of 1000 adults from the U.S.A. and 1000 adults from Germany were asked whether they supported the death penalty. 560 American adults and 590 German adults indicated that they supported the death penalty. Researcher wants to know if there is sufficient evidence to conclude that the proportion of adults who support the deal penally in the US is different than that in Germany. Use a confidence interval (at confidence interval of 98%) to...
Please type out answers. A recent survey showed that in a sample of 100 elementary school...
Please type out answers. A recent survey showed that in a sample of 100 elementary school teachers, 15 were single. In a sample of 180 high school teachers, 36 were single. Is the proportion of high school teachers who were single greater than the proportion of elementary teachers who were single? Use α = 0.01.a) State the hypotheses that need to be tested (using proper notation) b) Calculate the pooled estimate of p. c) Calculate the test statistic. d) Calculate...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT