Problem 1:
Medicare would like to test the hypothesis that the average monthly rate for one-bedroom assisted-living facility is equal to $3,300. A random sample of 12 assisted-living facilities had an average rate of $3,690 per month. The standard deviation for this sample was $530. Medicare would like to set α = 0.05. The correct hypothesis statement for this hypothesis test would be
H0: μ ≥ $3,300; H1: μ < $3,300. |
H0: μ = $3,300; H1: μ ≠ $3,300. |
H0: x̄ = $3,300; H1: x̄ ≠ $3,300. |
H0: μ ≠ $3,300; H1: μ = $3,300. |
Problem: 2
Medicare would like to test the hypothesis that the average monthly rate for one-bedroom assisted-living facility is equal to $3,300. A random sample of 12 assisted-living facilities had an average rate of $3,690 per month. The standard deviation for this sample was $530. Medicare would like to set α = 0.05. The critical value for this hypothesis test would be
1.985 |
2.201 |
2.761 |
1.645 |
Problem 3:
Medicare would like to test the hypothesis that the average monthly rate for one-bedroom assisted-living facility is equal to $3,300. A random sample of 12 assisted-living facilities had an average rate of $3,690 per month. The standard deviation for this sample was $530. Medicare would like to set α = 0.05. The conclusion for this hypothesis test would be that because the absolute value of the test statistic is
less than the absolute value of the critical value, we can conclude that the average monthly rate for an assisted-living facility is not equal to $3,300. |
more than the absolute value of the critical value, we cannot conclude that the average monthly rate for an assisted-living facility is not equal to $3,300. |
less than the absolute value of the critical value, we cannot conclude that the average monthly rate for an assisted-living facility is not equal to $3,300. |
more than the absolute value of the critical value, we can conclude that the average monthly rate for an assisted-living facility is not equal to $3,300. |
Problem 4:
Medicare would like to test the hypothesis that the average monthly rate for one-bedroom assisted-living facility is equal to $3,300. A random sample of 12 assisted-living facilities had an average rate of $3,690 per month. The standard deviation for this sample was $530. Medicare would like to set α = 0.05. The p-value for this hypothesis test would be between ________.
0.02 and 0.05 |
0.01 and 0.025 |
0.05 and 0.10 |
0.10 and 0.25 |
Problem 5:
Medicare would like to test the hypothesis that the average monthly rate for one-bedroom assisted-living facility is equal to $3,300. A random sample of 12 assisted-living facilities had an average rate of $3,690 per month. The standard deviation for this sample was $530. Medicare would like to set α = 0.05. Which one of the following statements is true?
Because the p-value is less than α, we reject the null hypothesis and conclude that the average monthly rate for an assisted-living facility is not equal to $3,300. |
Because the p-value is greater than α, we fail to reject the null hypothesis and conclude that the average monthly rate for an assisted-living facility is equal to $3,300. |
Because the p-value is greater than α, we fail to reject the null hypothesis and cannot conclude that the average monthly rate for an assisted-living facility is not equal to $3,300. |
Because the p-value is less than α, we fail to reject the null hypothesis and conclude that the average monthly rate for an assisted-living facility is not equal to $3,300. |
Problem 1: Medicare would like to test the hypothesis that the average monthly rate for one-bedroom...
Medicare would like to test the hypothesis that the average monthly rate for one-bedroom assisted-living facility is equal to $3,300. A random sample of 12 assisted-living facilities had an average rate of $3,690 per month. The standard deviation for this sample was $530. Medicare would like to set α = 0.05. The conclusion for this hypothesis test would be that because the absolute value of the test statistic is less than the absolute value of the critical value, we cannot...
Medicare would like to test the hypothesis that the average monthly rate for one-bedroom assisted-living facility is equal to $3,300. A random sample of 12 assisted-living facilities had an average rate of $3,690 per month. The standard deviation for this sample was $530. Medicare would like to set α = 0.05. The p-value for this hypothesis test would be between Options: 0.01 and 0.02 0.02 and 0.05 0.05 and 0.10 0.10 and 0.20
Medicare would like to test the hypothesis that the average monthly rate for one-bedroom assisted-living facility is equal to $3,300. A random sample of 12 assisted-living facilities had an average rate of $3,690 per month. The standard deviation for this sample was $530. Medicare would like to set α = 0.05. The test statistic for this hypothesis test would be ________. 1.98 2.55 -2.16 -1.37
Suppose that Medicare would like to test the hypothesis that the average monthly rate for one-bedroom assisted-living facility is equal to $3,300. A random sample of 12 assisted-living facilities had an average rate of $3.690 per month. The standard deviation for this sample was $530, Medicare would like to set ?= 0.05. What is the correct hypothesis statement for this hypothesis test? 0 A. Ho x-S3.300, H1 :#53300 ??. ?? ?#53,300, ??.?-33,300 O c. HOP 2 $3.300, HI : <...
9. Medicare would like to test the hypothesis that the average monthly rate for a one- bedroom assisted-living facility is equal to $3,300. A random sample of 12 assisted-living facilities had an average rate of $3,690 per month. The standard deviation for this sample was $530. Medicare would like to set a = 0.05. +$3,300 and it is a Note: Two hypotheses for the test are H :u=$3,300 vs H : two-tailed test. Calculate the test statistics • find the...
Problem 1: Expedia would like to test the hypothesis that the average roundtrip airfare between Philadelphia and Paris is higher for a flight originating in Philadelphia when compared to a flight originating in Paris. The following data summarizes the sample statistics for roundtrip flights originating in both cities. Assume that the population variances are equal. Originating City Philadelphia Paris Sample mean $1,240 $1,060 Sample size 15 19 Sample standard deviation $270 $240 If Population 1 is defined as flights originating...
Exam Pages Previous 1/2 2) A local Chamber of Commerce would like to test the hypothesis that the average monthly rent for (15p a one-bedroom apartment in Lakeland is different from the average rent of a one-bedroom apartment in Tampa. Assume that the population variances are equal and use alpha-0.05. In addition, calculate the 95% confidence interval and interpret. Tampa Lakeland Sample mean $830 $770 Sample size 1418 Sample standard deviation $75 $65 Verdana 9 Α Α BI 2 Ι....
A company would like to test the hypothesis that the average length of an online video watched by a user is more than 8 minutes. A random sample of 37 people watched online videos that averaged 8.7 minutes in length. It is believed that the population standard deviation for the length of online videos is 2.5 minutes. YouTube would like to set α = 0.02. The conclusion for this hypothesis test would be that because the test statistic is _____________________. more...
The Department of Education would like to test the hypothesis that the average debt load of graduating students with a bachelor's degree is equal to $17,000. A random sample of 34 students had an average debt load of $18,200. It is believed that the population standard deviation for student debt load is $4,200. The α is set to 0.05. The confidence interval for this hypothesis test would be ________.
The Department of Education would like to test the hypothesis that the average debt load of graduating students with a Bachelor's degree is equal to $17,000. A random sample of 34 students had an average debt load of $18,200. It is believed that the population standard deviation for student debt load is $4,200. The Department of Education would like to set α = 0.05. The critical sample means for this hypothesis test would be Question options: $14,118.9, $19,881.2 $14,839.1, $19,160.9...