Question

9. (19) A random sample of 64 UPW college students shows that the sample mean GPA is 2.82 with a standard deviation of 0.45.

(a) Construct a 90% Confidence Interval for the mean GPA of all UPW students.

(b) If we want to be 95% confident, and we want to control the maximum error of estimation to 0.1, how many more students should be added into the given sample?

(c) Would you conclude that the mean GPA in UPW is lower than 3 at 5% level of significance?

Answer #1

a) For 90% Confidence interval, the critical value is
z_{0.05} = 1.645

The 90% confidence interval is

= 2.82 +/- 0.09

= 2.73, 2.91

b) Margin of error = 0.1

c)

The test statistic is

P-value = P(Z < -3.2)

= 0.0007

Since the P-value is less than the significance level (0.0007 < 0.05), so we should reject the null hypothesis.

At 0.05 significance level, there is sufficient evidence to conclude that the mean GPA in UPW is lower than 3.

A random sample of 50 students has a mean GPA of 2.55. It is
known that the population standard deviation for the GPA of all
students is 1.1. Use this information to complete the
following:
a) Does the information above provide an estimate of a
population mean or a population proportion?
b) Provide a point estimate for the population mean GPA of all
students.
c) Construct and interpret a 95% confidence interval for the
parameter you selected in part (a)....

In a random sample of 64 audited income tax returns, it was
determined that the mean amount of additional tax owed was $3448
with a standard deviation of $2580. Construct and interpret a 90%
confidence interval for the mean additional amount of tax owed for
income tax-returns.
Lower bound: ___
Upper bound: ___
Interpret and choose correct answer:
A) One can be 90% confident that the mean additional tax owed is
between the lower and upper bounds
B) One can...

The mean age for all Foothill College students for a recent Fall
term was 33.2. The population standard deviation has been pretty
consistent at 15. Suppose that twenty-five Winter students were
randomly selected. The mean age for the sample was 30.4. We are
interested in the true mean age for Winter Foothill College
students. Let X = the age of a Winter Foothill College student.
A. Construct a 95% Confidence Interval for the true mean age of
Winter Foothill College...

A college admissions director wishes to estimate the mean age of
all students currently enrolled. In a random sample of 19 students,
the mean age is found to be 23.7 years. From past studies, the ages
of enrolled students are normally distributed with a standard
deviation of 10.4 years. Construct a 90% confidence interval for
the mean age of all students currently enrolled.
1. The critical value:
2. The standard deviation of the sample mean:
3. The margin of error...

In a random sample of 64 audited estate tax returns, it was
determined that the mean amount of additional tax owed was $3445
with a standard deviation of $2593. Construct and interpret a 90%
confidence interval for the mean additional amount of tax owed for
estate tax returns. LOADING... Click the icon to view the
t-distribution table. The lower bound is $ nothing. (Round to the
nearest dollar as needed.) The upper bound is $ nothing. (Round
to the nearest...

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 about 7 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
6.2 hours with a standard deviation of 0.97 hours. We want to
construct a 95% confidence interval for the mean nightly hours of...

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 about 7 hours. You randomly select 45
college students and survey them on their sleep habits. From this
sample, the mean number of hours of sleep is found to be 6.2 hours
with a standard deviation of 0.97 hours. We want to construct a 95%
confidence interval for the mean nightly hours of...

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 about 7 hours. You randomly select 45 college students
and survey them on their sleep habits. From this sample, the mean
number of hours of sleep is found to be 6.2 hours with a standard
deviation of 0.97 hours. We want to construct a 95% confidence
interval for the mean nightly hours of...

Test the claim that the mean GPA of night students is
significantly different than 2.9 at the 0.1 significance level.
Based on a sample of 20 people, the sample mean GPA was 2.87
with a standard deviation of 0.06
The test statistic is (to 3 decimals)
The positive critical value is (to 3
decimals)
Based on this we
fail to reject the null hypothesis
reject the null hypothesis

In a simple random sample of 500 freshman students, 276 of them
live on campus. You want to claim that majority of college freshman
students in US live on campus.
A) Are the assumptions for making a 95% confidence interval for
the true proportion of all college freshman students in the United
States who live on campus satisfied? yes/no?
B) What is the sample proportion?
C.) A 95% confidence interval for the actual proportion of
college freshman students in the...

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 3 minutes ago

asked 9 minutes ago

asked 12 minutes ago

asked 23 minutes ago

asked 34 minutes ago

asked 40 minutes ago

asked 48 minutes ago

asked 50 minutes ago

asked 51 minutes ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago