Question
In: Statistics and Probability
Random samples of female and male UVA undergraduates are asked to estimate the number of alcoholic...
Random samples of female and male UVA undergraduates are asked to estimate the number of alcoholic drinks that each consumes on a typical weekend. The data is below:
Females (Population 1): 2, 2, 2, 1, 3, 1, 3, 5, 5, 3
Males (Population 2): 5, 5, 6, 4, 4, 6, 3, 3, 3, 5
Give a 92.3% confidence interval for the difference between mean female and male drink consumption. (Assume that the population variances are equal.)
Confidence Interval =
Solutions
Expert Solution
| P1 | P2 | |
| 2 | 5 | |
| 2 | 5 | |
| 2 | 6 | |
| 1 | 4 | |
| 3 | 4 | |
| 1 | 6 | |
| 3 | 3 | |
| 5 | 3 | |
| 5 | 3 | |
| 3 | 5 | |
| Mean | 2.7 | 4.4 |
| SD | 1.418136492 | 1.173787791 |
Excel formulas:
mean = Average(B2:B11)
SD = STDEV(B2:B11)
orchestra answered 2 years agoRelated Solutions
Random samples of female and male UVA undergraduates are asked to estimate the number of alcoholic...
Random samples of female and male UVA undergraduates are asked
to estimate the number of alcoholic drinks that each consumes on a
typical weekend. The data is below:
Females (Population 1): 4, 5, 3, 6, 6, 4, 5, 3, 4, 2
Males (Population 2): 4, 6, 8, 4, 5, 6, 7, 6, 6, 7
Give a 95% confidence interval for the difference between mean
female and male drink consumption. For each confidence interval,
enter your answer in the form (LCL,...
An Office of Admissions document claims that 54.6% of UVA undergraduates are female. To test whether...
An Office of Admissions document claims that 54.6% of UVA
undergraduates are female. To test whether this claim is accurate,
a random sample of 225 UVA undergraduates was selected. In this
sample, 54.6667% were female. Is there sufficient evidence to
conclude that the document's claim is false? Carry out a hypothesis
test at a 1% significance level.
The p-value is:
Each person in random samples of 250 male and 250 female working American adults was asked...
Each person in random samples of 250 male and 250 female working
American adults was asked how long, in minutes, they typically
spend on their lunch break. The males averaged 39.6 minutes with a
standard deviation of 24.3 minutes. The females averaged 37.3
minutes with a standard deviation of 24.0 minutes. Is there
convincing evidence that the mean lunch break times for male and
female working American adults differ?
(a) State the null and alternative hypotheses to be tested.
(...
A random sample of 500 male and female adults was asked the amount of time each...
A random sample of 500 male and female adults was asked the
amount of time each person spent watching TV last week. Their
responses are shown at the right. At the 0.05 significance level,
does it appear that the amount of time spent watching TV is related
to the gender of the viewer?
Hours/Gender
Male
Female
Total
Under 8
90
110
200
8 to 15
85
75
160
15 or more
75
65
140
Total
250
250
500
Which test...
Each person in random samples of 227 male and 283 female working adults living in a...
Each person in random samples of 227 male and 283 female working
adults living in a certain town in Canada was asked how long, in
minutes, his or her typical daily commute was. (Use a statistical
computer package to calculate the P-value. Use
μmales − μfemales. Round
your test statistic to two decimal places, your df down to the
nearest whole number, and your P-value to three decimal
places.)
Males
Females
Sample
size
x
s
Sample
size
x
s
227...
Each person in random samples of 227 male and 253 female working adults living in a...
Each person in random samples of 227 male and 253 female working
adults living in a certain town in Canada was asked how long, in
minutes, his or her typical daily commute was. (Use a statistical
computer package to calculate the P-value. Use
μmales − μfemales. Round
your test statistic to two decimal places, your df down to the
nearest whole number, and your P-value to three decimal
places.)
Males
Females
Sample
size
x
s
Sample
size
x
s
227...
Independent random samples were taken of male and female members of University Entrepreneurship Club. These members...
Independent random samples were taken of male and female members
of University Entrepreneurship Club. These members were considering
starting a business. Of 500 members, 150 actually started a
business venture.
Estimate the value of the population proportion. (Round
your answers to 3 decimal places.)
Develop a 99% confidence interval for the population
proportion.
VERY URGENT !!!! Independent random samples were taken of male and female members of University Entrepreneurship...
VERY URGENT !!!! Independent random samples were taken of male
and female members of University Entrepreneurship Club. These
members were considering starting a business. Of 350 members, 250
actually started a business venture. (use 3 digits after
decimal point)
a. Estimate the value of the population
proportion. (Answer)
b. 90% confidence interval for the population
proportion is [Answer to Answer]
Researchers equipped random samples of 46 male and 58 female students from a large university with...
Researchers equipped random samples of 46 male and 58 female
students from a large university with a small device that secretly
records sound for a random 30 seconds during each 12.5-minute
period over two days. Then they counted the number of words spoken
by each subject during each recording period and, from this,
estimated how many words per day each subject speaks. The female
estimates had a mean of 16,177 words per day with a standard
deviation of 7520 words...
Two random samples are taken, one from among UVA students and the other from...
Two random samples are taken, one from among UVA students and the other from among UNC students. Both groups are asked if academics are their top priority. A summary of the sample sizes and proportions of each group answering yes" are given below: UVA (Pop. 1): n1 = 89,\(\hat{p}_{1}\)
= 0.81 UNC (Pop. 2): n2 = 86,\(\hat{p}_{2}\) = 0.561 Find a 97.3% confidence interval for the difference P1 – P2 of the population proportions. Confidence interval = _______
ADVERTISEMENT
ADVERTISEMENT
Latest Questions
- Compute the cost of not taking the following cash discounts. (Use a 360-day year. Do not...
- During his project presentation, Bob falls sick and performs poorly. However, his colleague, Dan, concludes that...
- A sample human input file that contains test values that a human would enter on the...
- A solution is prepared by dissolving 15 ml of methanol in 150.0 ml water at 25...
- Consider a newlywed who is planning a wedding anniversary gift of a trip to Canada for...
- We are evaluating a project that costs $868,000, has a life of eight years, and has...
- a.) Given sample data: 61.2, 61.9, 62.8, 63.1, 64.0, 64.3, 64.9, 65.5, 66.3 and 67.9, test...
ADVERTISEMENT