Respuesta :
Answer:
We conclude that the mean cost-to-charge ratio for Oregon hospitals is lower for outpatient care than for inpatient care.
Step-by-step explanation:W
We are given with the cost-to-charge ratios for both inpatient and outpatient care in 2002 for a sample of six hospitals in Oregon below;
Hospital 2002 Inpatient Ratio 2002 Outpatient Ratio
1 68 54
2 100 75
3 71 53
4 74 56
5 100 74
6 83 71
Let [tex]\mu_1[/tex] = mean cost-to-charge ratio for outpatient care
[tex]\mu_2[/tex] = mean cost-to-charge ratio for impatient care.
SO, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu_1 \geq \mu_2[/tex] {means that the mean cost-to-charge ratio for Oregon hospitals is higher or equal for outpatient care than for inpatient care}
Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu_1 < \mu_2[/tex] {means that the mean cost-to-charge ratio for Oregon hospitals is lower for outpatient care than for inpatient care}
The test statistics that would be used here is Two-sample t-test statistics because we don't know about population standard deviations;
T.S. = [tex]\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] ~ [tex]t__n__1_+_n_2_-_2[/tex]
where, [tex]\bar X_1[/tex] = sample mean cost-to-charge Outpatient Ratio = [tex]\frac{\sum X_1}{n_1}[/tex] = 63.83
[tex]\bar X_2[/tex] = sample mean cost-to-charge Impatient Ratio = [tex]\frac{\sum X_2}{n_2}[/tex] = 82.67
[tex]s_1[/tex] = sample standard deviation for Outpatient Ratio = [tex]\sqrt{\frac{\sum (X_1-\bar X_1 )^{2} }{n_1-1} }[/tex] = 10.53
[tex]s_2[/tex] = sample standard deviation for Impatient Ratio = [tex]\sqrt{\frac{\sum (X_2-\bar X_2 )^{2} }{n_2-1} }[/tex] = 14.33
[tex]n_1[/tex] = sample of hospital for outpatient care = 6
[tex]n_2[/tex] = sample of hospital for outpatient care = 6
Also, [tex]s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }[/tex] = [tex]\sqrt{\frac{(6-1)\times 10.53^{2}+(6-1)\times 14.33^{2} }{6+6-2} }[/tex] = 12.574
So, the test statistics = [tex]\frac{(63.83-82.67)-(0)}{12.574 \times \sqrt{\frac{1}{6}+\frac{1}{6} } }[/tex] ~ [tex]t_1_0[/tex]
= -2.595
The value of t test statistics is -2.595.
Now, at 5% significance level, the t table gives critical value of -1.812 at 10 degree of freedom for left-tailed test.
Since, our test statistics is less than the critical value of t as -2.595 < -1.812, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.
Therefore, we conclude that the mean cost-to-charge ratio for Oregon hospitals is lower for outpatient care than for inpatient care.
