Question

9. (19) A random sample of 64 UPW college students shows that the sample mean GPA...

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?

Homework Answers

Answer #1

a) For 90% Confidence interval, the critical value is z0.05 = 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.

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
A random sample of 50 students has a mean GPA of 2.55. It is known that...
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)....
1.The following data shows test scores for a sample of statistics students. 83, 64, 84, 76,...
1.The following data shows test scores for a sample of statistics students. 83, 64, 84, 76, 84, 54, 75, 59, 70, 63, 80, 84, 73, 68, 52, 65, 90, 52, 95, 36, 78, 61, 59, 84, 95, 47, 87 a. Find a 95% confidence interval for the mean test score for all students. b. Interpret this 95% CI for the population mean. c. What is the margin of error? 2. A recent national survey found that high school students watched...
In a random sample of 64 audited income tax returns, it was determined that the mean...
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...
A college admissions director wishes to estimate the mean age of all students currently enrolled. In...
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...
The mean age for all Foothill College students for a recent Fall term was 33.2. The...
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...
In a random sample of 64 audited estate tax​ returns, it was determined that the mean...
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...
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...
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...
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...
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
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT