The mean height of 50 male students who showed
above-average participation in college
athletics was 68.2 inches with a standard deviation of 2.5 inches,
while 50 male students
who showed no interest in such participation had a mean height of
67.5 inches with a
standard deviation of 2.8 inches.
(a) Test the hypothesis that male students who participate in
college athletics are taller
than other male students.
(b) What is the P value of the test?
In: Statistics and Probability
20) A research study investigated differences between male and female students. Based on the study results, we can assume the population mean and standard deviation for the GPA of male students are µ = 3.5 and σ = .05. Suppose a random sample of 100 male students is selected and the GPA for each student is calculated. Find the interval that contains 95.44 percent of the sample means for male students. µ = Population Mean σ =Population Standard Deviation
In: Statistics and Probability
Grade point averages of students on a large campus follow a normal distribution with a mean of 2.6 and a standard deviation of 0.5. d) A random sample of 400 students is chosen from this campus. what is the probability that at least 80 of these students have grade point averages higher than 3.0? e) Two students are chosen at random from this campus. what is the probability that at least one of them has a grade point average higher than 3.0?
In: Statistics and Probability
In: Statistics and Probability
A population of 1,000 students spends an average of $10.50 a day on dinner. The standard deviation of the expenditure is $3. A simple random sample of 64 students is taken.
a. What are the expected value, standard deviation, and shape of the sampling distribution of the sample mean?
b. What is the probability that these 64 students will spend an average of more than $11 per person?
c. What is the probability that these 64 students will spend an average between $10 and $11 per person?
In: Math
(1 point) Coaching companies claim that their courses can raise the SAT scores of high school students. But students who retake the SAT without paying for coaching also usually raise their scores. A random sample of students who took the SAT twice found 427 who were coached and 2733 who were uncoached. Starting with their verbal scores on the first and second tries, we have these summary statistics:
Try 1 Try 2 Gain
| nn | x¯¯¯x¯ | ss | x¯¯¯x¯ | ss | x¯¯¯x¯ | ss | |
| Coached | 427 | 500 | 92 | 529 | 97 | 29 | 59 |
| Uncoached | 2733 | 506 | 101 | 527 | 101 | 21 | 52 |
Estimate a 99% confidence interval for the mean gain of all students who are coached.
___________________ to
at 99% confidence.
Now test the hypothesis that the score gain for coached students is
greater than the score gain for uncoached students. Let μ1 be the
score gain for all coached students. Let μ2 be the score gain for
uncoached students.
(a) Give the alternative hypothesis: μ1−μ2_______________ 0
(b) Give the tt test statistic:
(c) Give the appropriate critical value for α=5%:
In: Statistics and Probability
Sleep – College Students: Suppose you perform a study about the hours of sleep that college students get. You know that for all people, the average is 7.0 hours with a standard deviation of 1.3 hours. You randomly select 50 college students and survey them on their sleep habits. From this sample, the mean number of hours of sleep is found to be 6.40 hours. Assume the population standard deviation for college students is the same as for all people. We want to construct a 95% confidence interval for the mean number of hours of sleep for all college students.
(a) What is the point estimate for the mean amount of sleep for
all college students?
hours
(b) What is the critical value of z (denoted
zα/2) for a 95% confidence interval?
Use the value from the table or, if using software, round
to 2 decimal places.
zα/2 =
(c) What is the margin of error (E) for the mean number of
hours of sleep for all college students for a 95% confidence
interval? Round your answer to 3 decimal
places.
E = hours
(d) Construct the 95% confidence interval for the mean number of
hours of sleep for all college students. Round your answers
to 2 decimal places.
< μ <
In: Statistics and Probability
A college research group reported that 43% of college students aged 18-24 would spend their spring breaks relaxing at home in 2009. A sample of 165 college students was selected. Complete parts a through d below.
a. Calculate the standard error of the proportion.
σp = (Round to four decimal places as needed.)
b. What is the probability that less than 40% of the college students from the sample spent their spring breaks relaxing at home?
P(Less than 40% of the college students from the sample spent their spring breaks relaxing at home) = (Round to four decimal places as needed.)
c. What is the probability that more than 50% of the college students from the sample spent their spring breaks relaxing at home?
P(More than 50% of the college students from the sample spent their spring breaks relaxing at home) = (Round to four decimal places as needed.)
d. What is the probability that between 44% and 54% of the college students from the sample spent their spring breaks relaxing at home?
P(Between 44% and 54% of the college students from the sample spent their spring breaks relaxing at home) = (Round to four decimal places as needed.)
In: Statistics and Probability
Harold Jones, the financial aid officer at a small university, manages all aspects of the financial aid program. Jones receives requests for aid from students, determines whether the students meet the aid criteria, authorizes aid payments, notifies the applicants that their request has been either approved or denied, writes the financial aid checks on the account he controls, and requires that the students come to his office to receive the checks in person. For years, Jones has used his position of authority to perpetrate the following fraud.
Jones encourages students who clearly will not qualify to apply for financial aid. Although the students do not expect aid, they apply on the off chancethat it will be awarded. Jones modifies the financial information in the students’ applications so that it falls within the established guidelines for aid. He then approves aid and writes aid checks payable to the students. The students, however, are informed that aid was denied.
Because the students expect no aid, the checks in Jones’s office are never collected. Jones forges the students’ signatures and cashes the checks.
Required
Identify the internal control procedures (classified per COSO) that could prevent or detect this fraud.
In: Computer Science
Calculate the cell potential for a reaction in a electrolytic cell with the following half-reactions if: [U 3+] = 0.10 M, [MnO4 - ] = 0.20M, [Mn2+], and [H+ ] = 0.20 M
U 3+ + 3e -> U o Ecell = - 1.642 V
MnO4 - + 8H+ + 5e- -> Mn2+ + 4H2O Ecell = +1.51 V
In: Chemistry