Question

In a recent sample of $84$used car sales costs, the sample mean was $6,425$with a standard deviation of $3,156$. Assume the underlying distribution is approximately normal.

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

b. Define the random variable 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. For this problem we should use the Student's t-distribution with $\mathit{n}\mathbf{-}\mathbf{1}\mathbf{=}\mathbf{83}$the degrees of freedom, where $n$represents the size of the sample. b. Random variable $\overline{X}$is the mean number of costs from a sample of $84$used car sales. c. A $95\%$confidence interval for the population mean is$5.736\le \mu \le 7.114$ and E B M=$0.689$. d. With $95\%$confidence the true population mean is between $5.736$and $7.114$.

Step-by-step solution

Expression of variable random (part a)

For this problem we should use the Student's t-distribution with $n-1=83$$n-1=83$the degrees of freedom, where $n$represents the size of the sample. We use the Student's t-distribution because we don't know the standard deviation for the population. Random variable $\overline{X}$is the mean number of costs from a sample of $84$ used car sales. If $\overline{x}$and $\mathrm{S}$are the mean and standard deviation of a random sample from a normal distribution with unknown variance ${\sigma }^2,\$\backslash \text{textbf}\left\{\text{a}100\right(1-\alpha )\%$confidence interval on$\mu \}$ is given by

$\overline{x}-t_{\frac{\alpha }{2},n-1}\frac{s}{\sqrt{n}}\le \mu \le \overline{x}+t_{\frac{\alpha }{2},n-1}\frac{s}{\sqrt{n}}$

where $t\frac{\alpha }{2},n-1$is the upper $100\frac{\alpha }{2}$percentage point of the $t$distribution with $n-1$degrees of freedom.

We need to find a $95\%$on the population mean, then

$\frac{\alpha }{2}=\frac{1-0.95}{2}=0.025$$t_{\frac{\alpha }{2},n-1}=t_{0.025,83}=1.99$

We used a probability table for the student 's $t$- distribution to find the value of $t$. The table gives t-scores that correspond to degrees of freedom (row) and the confidence level (column). The t-score is found where the row and column intersect in the table.

We know that the sample mean was with a standard deviation of s = 3.156

Now from the equation (1) and (2) we get

Sketch the graph

• A $95\%$CI for$\mu$is

$5.736\le \mu \le 7.114$

• The graph of this problem is

Expand of variable 

• The error bound is

$EBM$$=$$0.689.$

$4257b2${ with $95\%$confidence the true population mean is between $5.736$and $7.114$}

• Ninety - five percent of all confidence intervals constructed in this way contain the true mean statistics exam score. For example, if we constructed $100$of these confidence intervals, we would expect$95$of them to contain the true population mean.