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Usimov [2.4K]
3 years ago
8

The expression 475 * 1.076 ^ t the average annual per capita health care costs, in dollars, in the US as a function of the numbe

r of years since 1970. What does 1.076 represent in this expression?
Mathematics
2 answers:
Alik [6]3 years ago
7 0

Answer:

Y(t)= 475 (1.076)^t

Where Y(t) represent the average annual per capita health care costs

475 represent the initial amount for the average annual per capita health care costs

t represent the number of years since 1970

And 1.076 represent the growth factor given by:

1+r = 1.076

And solving for r we got:

r = 1.076-1 =0.076

So for this case we can say that the value 1.076 represent the growth factor.

Step-by-step explanation:

For this case we have the following model given:

Y(t)= 475 (1.076)^t

Where Y(t) represent the average annual per capita health care costs

475 represent the initial amount for the average annual per capita health care costs

t represent the number of years since 1970

And 1.076 represent the growth factor given by:

1+r = 1.076

And solving for r we got:

r = 1.076-1 =0.076

So for this case we can say that the value 1.076 represent the growth factor.

Whitepunk [10]3 years ago
4 0

Answer:

the correct answer is C

Step-by-step explanation:

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1 year ago
At Western University the historical mean of scholarship examination scores for freshman applications is 900. A historical popul
vampirchik [111]

Answer:

a) Null Hypothesis: \mu =900

Alternative hypothesis: \mu \neq 900

b) The 95% confidence interval would be given by (910.05;959.95)    

c) Since we confidence interval not ocntains the value of 900 we fail to reject the null hypothesis that the true mean is 900.

d) z=\frac{935 -900}{\frac{180}{\sqrt{200}}}=2.750

Since is a bilateral test the p value is given by:

p_v =2*P(Z>2.750)=0.0059

Step-by-step explanation:

a. State the hypotheses.

On this case we want to check the following system of hypothesis:

Null Hypothesis: \mu =900

Alternative hypothesis: \mu \neq 900

b. What is the 95% confidence interval estimate of the population mean examination  score if a sample of 200 applications provided a sample mean x¯¯¯= 935?

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X=935 represent the sample mean for the sample  

\mu population mean (variable of interest)

\sigma=180 represent the population standard deviation

n=200 represent the sample size  

The confidence interval for the mean is given by the following formula:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}   (1)

In order to calculate the mean and the sample deviation we can use the following formulas:  

\bar X= \sum_{i=1}^n \frac{x_i}{n} (2)  

s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}} (3)  

The mean calculated for this case is \bar X=3278.222

The sample deviation calculated s=97.054

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that z_{\alpha/2}=1.96

Now we have everything in order to replace into formula (1):

935-1.96\frac{180}{\sqrt{200}}=910.05    

935+1.96\frac{180}{\sqrt{200}}=959.95    

So on this case the 95% confidence interval would be given by (910.05;959.95)    

c. Use the confidence interval to conduct a hypothesis test. Using α= .05, what is your  conclusion?

Since we confidence interval not ocntains the value of 900 we fail to reject the null hypothesis that the true mean is 900.

d. What is the p-value?

The statistic is given by:

z=\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

If we replace we got:

z=\frac{935 -900}{\frac{180}{\sqrt{200}}}=2.750

Since is a bilateral test the p value is given by:

p_v =2*P(Z>2.750)=0.0059

So then since the p value is less than the significance we can reject the null hypothesis at 5% of significance.

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