Question

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 value of the test statistic is a -0.05. The conclusion for this hypothesis test would be that because the absolute
A) 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
B) 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.
C) 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.
D) 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.

Answer 1

Answer:

C) 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.

Step-by-step explanation:

Hello!

Interest hypothesizes is " the average monthly rate for one-bedroom assisted-living facility is equal to $3300" symbolically: μ = 3300

The study variable is:

X: Monthly rate for a one-bedroom assisted-living facility.

Since there is no information about the distribution of the variable, to be able to study the population mean, I'll assume that the variable has a normal distribution.

The hypothesis is:

H₀: μ = 3300

H₁: μ ≠ 3300

α: 0.05

The statistic to use, considering that there is no known population information and the sample size, is a Student t:

t= X[bar] - μ ~$t_{n-1}$

       S/√n

n= 12

X[bar]= $3690

S= $530

Using the sample data, calculate the statistic value:

t= 3690 - 3300 = 2.549

       530/√12

The rejection region for this test is two-tailed, with critical values:

$t_{n-1; \alpha /2} = t_{11; 0.025} = -2.201$

$t_{n-1;1-\alpha /2} = t_{11; 0.975} = 2.201$

The decision rule is:

Reject the null hypothesis if t ≤ -2.201 or if t ≥ 2.201

Not reject the null hypothesis if -2.201 < t < 2.201

Since the calculated value (2.549) is greater than the right critical value (2.201) the decision is to reject the null hypothesis.

With a signification level of 5%, there is enough evidence to reject the null hypothesis. This means that the population mean of the monthly rate for a one-bedroom assisted living facility is different from $3300.

I hope it helps!