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 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

Answer 1

Null hypothesis (Ho) : The average monthly rate for one bedroom assisted living facility is $3300

$\mu = 3300$

Alternative hypothesis (H1) : The average monthly rate for one bedroom assisted living facility is not $3300.

H + 3300

Test statistic is given by -

$t = \frac{(\bar{x} - \mu_0)}{s/\sqrt{n}}$

where, is the sample mean = $3690

T

n is the sample size = 12

s is the sample standard deviation = $530

is the specified value of Population average under the null hypothesis = $3300

HO

Hence, the value of the test statistic will be -

$t = \frac{(3690-3300)}{530/\sqrt{12}}$

  $= \frac{390}{152.9}$

= 2.55

Degrees of freedom = n - 1 = 12 - 1 = 11

P value = P(|t| > 2.55)

= Area under the t curve with 11 degrees of freedom on the left side of t = -2.55 and right side of t = 2.55

= 0.027

(It can be obtained using the formula =TDIST(2.55,11,2) in excel).

So, P value will lie between 0.02 and 0.05