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Nimfa-mama [501]
3 years ago
5

Will give brainliest for thiss

Mathematics
2 answers:
anyanavicka [17]3 years ago
8 0

A:

4y + 3 = 19

Y = 3

B:

n = 16

Please mark as brainliest!

aleksandrvk [35]3 years ago
7 0
A. 4y + 3 = 19
3+4y=19
3 + -3 + 4y = 19 + -3
3 + -3 = 0
0 + 4y = 19 + -3
4y= 19 + -3
19 + -3 = 16
4y = 16
y = 4
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The Oregon Department of Health web site provides information on the cost-to-charge ratio (the percentage of billed charges that
Alekssandra [29.7K]

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 \mu_1 = <u><em>mean cost-to-charge ratio for outpatient care</em></u>

\mu_2 = <u><em>mean cost-to-charge ratio for impatient care</em></u>.

SO, Null Hypothesis, H_0 : \mu_1 \geq \mu_2     {means that the mean cost-to-charge ratio for Oregon hospitals is higher or equal for outpatient care than for inpatient care}

Alternate Hypothesis, H_A : \mu_1 < \mu_2     {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 <u>Two-sample t-test statistics</u> because we don't know about population standard deviations;

                         T.S. =  \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2}  } }  ~ t__n__1_+_n_2_-_2

where, \bar X_1 = sample mean cost-to-charge Outpatient Ratio = \frac{\sum X_1}{n_1} = 63.83

\bar X_2 = sample mean cost-to-charge Impatient Ratio = \frac{\sum X_2}{n_2} = 82.67

s_1 = sample standard deviation for Outpatient Ratio = \sqrt{\frac{\sum (X_1-\bar X_1 )^{2} }{n_1-1} } = 10.53

s_2 = sample standard deviation for Impatient Ratio = \sqrt{\frac{\sum (X_2-\bar X_2 )^{2} }{n_2-1} } = 14.33

n_1 = sample of hospital for outpatient care = 6

n_2 = sample of hospital for outpatient care = 6

Also, s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2}  }{n_1+n_2-2} } =  \sqrt{\frac{(6-1)\times 10.53^{2}+(6-1)\times 14.33^{2}  }{6+6-2} } = 12.574

So, <u><em>the test statistics</em></u>  =  \frac{(63.83-82.67)-(0)}{12.574 \times \sqrt{\frac{1}{6}+\frac{1}{6}  } }  ~ t_1_0

                                     =  -2.595

The value of t test statistics is -2.595.

<u>Now, at 5% significance level, the t table gives critical value of -1.812 at 10 degree of freedom for left-tailed test.</u>

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 <u><em>we reject our null hypothesis</em></u>.

Therefore, we conclude that the mean cost-to-charge ratio for Oregon hospitals is lower for outpatient care than for inpatient care.

8 0
3 years ago
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