Question

Medicare spending per patient in different U.S. metropolitan areas may differ. Based on the sample data below, answer the questions that follow to determine whether the average spending in the northern region is significantly less than the average spending in the southern region at the 1 percent level.

	Northern Region	Southern Region
Sample Mean	$4,327	$6865
Standard Deviation	1476	3602
Sample SIze	11	15
Significance	1%

(a-1) Choose the appropriate hypotheses. Assume μN is the average spending in the northern region and μS is the average spending in the southern region.

a. H0: μN − μS ≤ 0 versus H1: μN – μS >0

or

b. H0: μN − μS ≥ 0 versus H1: μN – μS < 0

(a-2) Specify the decision rule. (Do not use the quick rule to determine degrees of freedom. Round your answer to 3 decimal places. A negative value should be indicated by a minus sign.)

Reject the null hypothesis if tcalc < _________

(b) Find the test statistic tcalc assuming unequal variances. (Round your answer to 2 decimal places. A negative value should be indicated by a minus sign.)

tcalc -2.46

(c-1) Can we reject the null hypothesis?

No, fail to reject the null hypothesis.

or

Yes, reject the null hypothesis.

(c-2) The average spending in the northern region is significantly less than the average spending in the southern region.

True/False?

Answer 1

a-1) Hypothesis :

i.e, $H_0 : \mu _N - \mu _S \geq 0$

$H_1 : \mu _N < \mu _S$ i.e $H_0 : \mu _N - \mu _S < 0$

Left tailed test.

a-2)

Significance level = 1% = 0.01

degrees of freedom is given by,

$df =\frac{ \left [ \frac{s^2_1}{n_1} + \frac{s^2_2}{n_2} \right ]^2}{\frac{\left ( \frac{s^2_1}{n_1} \right )^2}{n_1 -1}+\frac{\left ( \frac{s^2_2}{n_2} \right )^2}{n_2 -1}}$

We are given , s1 = 1476   , s2 = 3602 , n1 = 11 , n2 = 15

$df =\frac{ \left [ \frac{1476^2}{11} + \frac{3602^2}{15} \right ]^2}{\frac{\left ( \frac{1476^2}{11} \right )^2}{11 -1}+\frac{\left ( \frac{3602^2}{15} \right )^2}{15 -1}}$

So , critical value for this left tailed test is,

$t_{\alpha,df } = t_{0.01,19.69931} =-2.539$

{ using Excel function ,   =T.INV( 0.01 , 19.69931 ) = -2.539 }

Reject the null hypothesis if t_calc < -2.539

b )

Test statistic :

$t_{calc} = \frac{\bar{x}_1 -\bar{x}_2 }{\sqrt{\frac{s^2_1 }{n_1}+\frac{s^2_2 }{n_2}}}$

$t_{calc} = \frac{4327 -6865 }{\sqrt{\frac{1476^2}{11}+\frac{3602^2 }{15}}}$

C-1 )

The rejection region for this left-tailed test is R = { t : t < - 2.539 }

It is observed that t_calc > -2.539

So, fail to reject the null hypothesis.

C-2)

false.

There is not enough evidence say that the average spending in the northern region is significantly less than the average spending in the southern region