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densk [106]
2 years ago
6

You are interested in purchasing a new car and have done some research. One of the many points you wish to consider is the resal

e value of the car after 5 years. You wish to estimate the resale value of the one that piques your interest with a 99% confidence interval. In your research, you obtain data on 17 recently resold 5-year-old sedans of the same model you wish to purchase. These 17 cars were resold at an average price of $12,500 with a standard deviation of $700. What is the 99% CI for the true mean resale value?
Business
1 answer:
kari74 [83]2 years ago
8 0

Answer:

The 99% confidence interval would be given by (12004.26;12995.74)  

Explanation:

1) Previous concepts

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=12500 represent the sample mean for the sample  

\mu population mean (variable of interest)

s=700 represent the sample standard deviation

n=17 represent the sample size  

2) Confidence interval

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

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=17-1=16

Since the Confidence is 0.99 or 99%, the value of \alpha=0.01 and \alpha/2 =0.005, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,16)".And we see that t_{\alpha/2}=2.92

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

12500-2.92\frac{700}{\sqrt{17}}=12004.26    

12500+2.92\frac{700}{\sqrt{17}}=12995.74    

So on this case the 99% confidence interval would be given by (12004.26;12995.74)    

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