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Problem 1: Medicare would like to test the hypothesis that the average monthly rate for one-bedroom...

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.
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Answer #1

Ans Problem -1 conect ophon is e Because, Medicare want to defeamine whe thes avesage mon thly vate is equ to zzoo o not they ave not intedested in wheheo it is less tan o geater than 300. Sovit is non-draechonal o two talled test. Pooblem-2 The coitical foo thts hypothesis tet wodd valve be 2. 201 That test is troo tailed Since populaton standard deviahon s unknocon so the t- tet caihca value toil be usedPooblem- 3 Correct opon is d The conclu sion o this hypothesls tet csdd be thot because the absolute value of the test statistic is mse thap the abgoute value of the coih cal moe than the abeoute valve of the coilt cal han the aksduie value coitfica VaJue, e can conclude that the avesage monthly vate fo ah assisted luna faciity is not equal to $3,300Pooblen-4 The corredl option ic a e The p-value fo this hypothesic tect wodld be between o.o2 and o.05 since fa df table volue of toos 2.201 and toon-2-7)g! レシto.or<2ST<t0.02. ooblem-5 The cortel option a ie Beauce the p-value is less than we oeječl the null hypothesis and conclude thaf -the average motnly vate f an asicied-ing facibty is not equel to 3,300

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