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Alexeev081 [22]

3 years ago

14

Myra is evaluating the expression –31.7 + 4.5x, when x = 2.1. –31.7 + 4.5(2.1) = –27.2(2.1) = –57.12 What was Myra’s error? A.My

ra should have multiplied -31.7 and 2.1 first. B. Myra should have multiplied 4.5 and 2.1 first. C. Myra should have subtracted -27.2 and 2.1 first. D. Myra should have added -31.7 and 2.1 first.

2 answers:

Art [367]3 years ago

8 0

B.myra should have multiplied 4.5 and 2.5 first.

melomori [17]3 years ago

4 0

Answer:

The answer is B.

Step-by-step explanation:

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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.

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