I am having a bit of trouble with the top half of the questions. I have an answer below for questions 5-10, is there anyway you can check those answers and help me with the rest I am having trouble with. Thank you.

1. The area to the right of 1.93 under Student's t-curve with 11 degrees of freedom is
a. Between 5% and 2.5%
b. More than 95%
c. Between 25% and 10%
d. Less than 1%

2. The area to the left of -1.5 under Student's curve with 9 degrees of freedom is
a. Between 25% and 10%
b. Between 10% and 5%
c. Exactly 7.5%
d. Between 95% and 90%

3. The area to the left of 1.25 under Student's t-curve with 20 degrees of freedom is
a. Less than 10%
b. Between 25% and 10%
c. More than 75%
d. More than 90%

4. You should use a t-test when which of the following apply? (Choose all of the correct responses.)
a. The size of the sample (draws from the box) is less than or equal to 25.
b. Only for hypothesis tests about the average.
c. For hypothesis tests about either the percentage or the average.
d. For every hypothesis test where you know the standard deviation of the box.

Use the information below to answer questions 5-7.
Commuter students at a certain college claim that the average distance they have to commute to campus is less than 26 miles per day. A random sample of 16 commuter students was surveyed and yielded an average distance of 29 miles and a SD of 8 miles. We want to test the hypothesis that the students' claim is correct.

5. The test statistic for this hypothesis test is closest to:
a. 0.5
b. 1.0
c. 1.5
d. 3

I think it is C

6. The p-value for this hypothesis is:
a. between 25% and 10%
b. between 10% and 5%
c. between 5% and 2.5%
d. between 1% and 0.5%

I think it is B

7. Based on the p-value, you can conclude that:
a. It is not plausible that the average commute is 29 miles per day.
b. It is plausible that the average commute is 29 miles per day.
c. It is not plausible that the average commute is less than 26 miles per day.
d. It is plausible that the average commute is less than 26 miles per day.

I think it is D

Use the information below to answer questions 8-10.
An advertising agency would like to create an advertisement for a fast-food restaurant claiming that the average waiting time from ordering to receiving your order at the restaurant is less than 5 minutes. The agency measured the time from ordering to deliver of order for 25 customers and found that the average time was 4.7 minutes with an SD+ of .6 minutes. Test the following hypotheses: Null: Waiting time is 5 minutes or more Alternative: Waiting time is less than 5 minutes

8. The test statistic is closest to:
a. -0.5
b. -2.5
c. 3.03
d. 12.5

I think it is B

9. The p-value is:
a. less than 1%
b. between 10% and 5%
c. between 5% and 2.5%
d. more than 95%

I think it is A

10. The appropriate conclusion is:
a. Reject the null hypothesis
b. Do not reject the null hypothesis
c. Reject the alternative hypothesis
d. Do not reject the alternative hypothesis

I think it is D

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Let's go through the questions step by step and check your answers.

1. The area to the right of 1.93 under Student's t-curve with 11 degrees of freedom is requesting the right-tail area. To find this, you can use a t-distribution table or a statistical software. Looking up the value in the t-distribution table, the area is less than 5%, which means the correct answer is d. Less than 1%.

2. The area to the left of -1.5 under Student's t-curve with 9 degrees of freedom is requesting the left-tail area. Again, you can use a t-distribution table or a statistical software to find this. Looking up the value in the t-distribution table, the area is less than 10% but greater than 5%, which means the correct answer is b. Between 10% and 5%.

3. The area to the left of 1.25 under Student's t-curve with 20 degrees of freedom is requesting the left-tail area. Using a t-distribution table or a statistical software, the area is less than 25% but greater than 10%, which means the correct answer is b. Between 25% and 10%.

4. You should use a t-test when which of the following apply? The correct answers are a. The size of the sample (draws from the box) is less than or equal to 25 and c. For hypothesis tests about either the percentage or the average. The t-test is specifically designed for smaller sample sizes and for hypothesis tests involving averages. The correct choices are a and c.

5. The test statistic for this hypothesis test is calculated by subtracting the hypothesized population mean (26) from the sample mean (29) and dividing it by the standard deviation of the sample mean (8 divided by the square root of 16). This gives us a test statistic of 3. The closest option is d. 3.

6. The p-value for this hypothesis test is the probability of observing a test statistic as extreme as or more extreme than the one obtained under the null hypothesis. With a test statistic of 3, the p-value is less than 5% but greater than 2.5%, so the correct answer is c. Between 5% and 2.5%.

7. Based on the p-value, we compare it to our predetermined significance level (commonly set at 5%). If the p-value is less than the significance level, we reject the null hypothesis. Since the p-value is between 5% and 2.5%, which is less than 5%, we can conclude that it is plausible that the average commute is less than 26 miles per day. The correct answer is d. It is plausible that the average commute is less than 26 miles per day.

8. The test statistic is calculated by subtracting the hypothesized population mean (5) from the sample mean (4.7) and dividing it by the standard deviation of the sample mean (0.6 divided by the square root of 25). This gives us a test statistic of -2.5. Therefore, the correct answer is b. -2.5.

9. The p-value for this hypothesis test is the probability of observing a test statistic as extreme as or more extreme than the one obtained under the null hypothesis. With a test statistic of -2.5, the p-value is less than 1%. So, the correct answer is a. less than 1%.

10. The appropriate conclusion is to compare the p-value to the predetermined significance level (commonly set at 5%). If the p-value is less than the significance level, we reject the null hypothesis. In this case, the p-value is less than 1%, which is less than 5%. Therefore, we can reject the null hypothesis and conclude that the average waiting time is less than 5 minutes. The correct answer is a. Reject the null hypothesis.

I hope this helps! Let me know if you have any further questions.

Let's go through each question and check your answers.

1. The area to the right of 1.93 under Student's t-curve with 11 degrees of freedom can be found using a t-table or a calculator with t-distribution functions. Looking up the t-value of 1.93 with 11 degrees of freedom, we find that the area to the right is between 5% and 2.5%. Therefore, the correct answer is a. Between 5% and 2.5%.

2. The area to the left of -1.5 under the Student's t-curve with 9 degrees of freedom can also be found using a t-table or a calculator. However, since we are looking for the area to the left of a negative value, we need to consider the absolute value of -1.5, which is 1.5. Looking up the t-value of 1.5 with 9 degrees of freedom, we find that the area to the left is between 5% and 2.5%. Therefore, the correct answer is c. Exactly 7.5%.

3. Again, we need to find the area to the left of 1.25 under the Student's t-curve with 20 degrees of freedom. Looking up the t-value of 1.25 with 20 degrees of freedom, we find that the area to the left is less than 10%. Therefore, the correct answer is a. Less than 10%.

4. The appropriate scenarios to use a t-test are when the sample size is small (less than or equal to 25), and when we are conducting hypothesis tests about the average. Therefore, the correct answers are a. The size of the sample (draws from the box) is less than or equal to 25, and b. Only for hypothesis tests about the average.

Now, let's check your answers for questions 5-10 on hypothesis testing.

5. The test statistic for this hypothesis test can be found by subtracting the null hypothesis value (26 miles) from the sample mean (29 miles), and then dividing by the standard deviation of the sample mean (8 miles divided by the square root of 16). This calculation gives us a test statistic of 0.75, which is closest to option a. 0.5. Therefore, the correct answer is a.

6. The p-value for this hypothesis test is the probability of observing a test statistic as extreme as, or more extreme than, the observed test statistic (0.75), assuming the null hypothesis is true. Since our test is a one-tailed test (less than), we need to find the area to the left of 0.75 under the t-distribution with 15 degrees of freedom (16 - 1). Using a t-table or calculator, we find that the area is between 10% and 5%. Therefore, the correct answer is b. Between 10% and 5%.

7. Based on the p-value, we compare it to our desired significance level (commonly 0.05 or 5%). If the p-value is less than the significance level, we reject the null hypothesis. In this case, the p-value is between 10% and 5%, which is greater than 5%. Therefore, we fail to reject the null hypothesis. The correct answer is b. It is plausible that the average commute is 29 miles per day.

Moving on to questions 8-10:

8. To find the test statistic, we calculate the difference between the sample mean (4.7 minutes) and the null hypothesis value (5 minutes), then divide by the standard deviation of the sample mean (0.6 minutes divided by the square root of 25). This calculation gives us a test statistic of -2.5, which is closest to option b. -2.5. Therefore, the correct answer is b.

9. To find the p-value, we need to find the area to the left of -2.5 under the t-distribution with 24 degrees of freedom (25 - 1). Using a t-table or calculator, we find that the p-value is less than 1%. Therefore, the correct answer is a. less than 1%.

10. Based on the p-value being less than our significance level of 0.05 (5%), we can conclude that we reject the null hypothesis. Therefore, the correct answer is a. Reject the null hypothesis.

I hope this helps! If you have any further questions, feel free to ask.