Question

In a recent sample of $84$used car sales costs, the sample mean was $\backslash (6,425$with a standard deviation of $\backslash )3,156$. Assume

the underlying distribution is approximately normal.

a. Which distribution should you use for this problem? Explain your choice.

b. Define the random variable$\overline{X}$in words.

c. Construct a $95\%$confidence interval for the population mean cost of a used car.

i. State the confidence interval.

ii. Sketch the graph.

iii. Calculate the error bound.

d. Explain what a “$95\%$ confidence interval” means for this study.

Short answer

(a) Because we don't know the standard deviation population, we can utilise a student's t distribution. The probability distribution is $t83.$

(b) The average cost of an 84-year-old automobile is $\overline{X}$.

(c) The final result we get,

1. $CI=(5740.1,7109.9)$

2. The following graph is:

3.E B M=665.0178

(d) If we sampled $n=84$many times, we would expect to see that $95\%$of the time, the genuine population mean selling price of a used car would be found.

Step-by-step solution

Explanation (a)

Because we don't know the standard deviation population, we can utilise a student's t distribution. The probability distribution is $t83.$

Explanation (b)

The average cost of an 84-year-old automobile is $\overline{X}$.

Explanation (c)

i. Using the $T I-83$ calculator, calculate the confidence interval.

Calculators TI-83, TI-83+, TI-84, TI-84+

Make a list of the information.

Press the STAT button and arrow $\to$button to TESTS.

Now, Arrow $\downarrow$to $8:$T Interval.

Press ENTER button also.

Arrow to $\to$and then press ENTER.

Arrow $\downarrow$and the name of the list in which the data is kept.

Enter Freq : 1 Enter $\mathit{C}$- Level : $.9$

Arrow $\downarrow$to Calculate and press Enter.

The confidence interval's output,

$CI=(5740.1,7109.9)$

ii. The graph is as follows:

iii. The error bound is calculated from the formula,

$EBM=t_{n-1}\left(\frac{a}{2}\right)\frac{s}{\sqrt{n}}$

$EBM=t_{84-1}\left(\frac{0.05}{2}\right)\frac{3156}{\sqrt{84}}$

$EBM=665.0178$

Explanation (d)

If we sampled $n=84$many times, we would expect to see that$95\%$ of the time, the genuine population mean selling price of a used car would be found.