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You are interested in purchasing a new car. One of the many points you wish to consider is the resale value of the car after 5 years of ownership. Since you are particularly interested in the Toyota Camry, you decide to estimate the resale value of the Camry with a 90% confidence interval. You manage to randomly obtain data on 17 recently resold 5 year old Toyota Camrys. These 17 cars were resold at an average price of $12,400. Assume the resale value of the Camry after 5 years of ownership is normally distributed.
Write the formula that should be used to calculate the 90% confidence interval for the true mean resale value of the Camry after 5 years of ownership.
If the 90% confidence interval for the true mean resale value of the Camry after 5 years of ownership is from $12,103 to $12,697, what is the value for the sample standard deviation of the 17 cars in the sample? Assume all values below are rounded to the nearest dollar.
a) $625
b) $701
c) $744
d) Can't answer the question based on the information given.
e) None of the above are correct.

Sagot :

Answer:

b) $701

Step-by-step explanation:

We want to find the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 17 - 1 = 16

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 16 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.9}{2} = 0.95[/tex]. So we have T = 1.746.

If the 90% confidence interval for the true mean resale value of the Camry after 5 years of ownership is from $12,103 to $12,697.

The margin of error is half this difference. So

[tex]M = \frac{12697-12103}{2} = 297[/tex]

Margin of error formula:

[tex]M = T\frac{s}{\sqrt{n}}[/tex]

In which s is the standard deviation of the sample and n is the size of the sample.

In this question:

[tex]T = 1.746, M = 297, n = 17[/tex]

We use this to find s, which is the sample standard deviation. So

[tex]M = T\frac{s}{\sqrt{n}}[/tex]

[tex]297 = 1.746\frac{s}{\sqrt{17}}[/tex]

[tex]s = \frac{297*\sqrt{17}}{1.746}[/tex]

[tex]s = 701.35[/tex]

To the nearest dollar, $701 and the answer is given by option b.

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