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 always use the Student's $t$-distribution with $n-1=83$the degrees of freedom, where $n$ represents the dimensions of the sample.

b. chance 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 $EBM=0.689$.

graph is

d. With $95\%$confidence actuality population mean is between$5.736$and$7.114$

Step-by-step solution

Statistical Power (a) and (b)

a) We must take the Independent samples $t$using $n-1=83$ variables for this example , where $n$ is de facto the big sample. once we wouldn't determine the majority's variance, we employ the Independent samples $t$. The common amount of charges from either a range of$84$ used vehicles deals is reflected by the statistical distribution $\overline{X}$. If $\overline{x}$and $s$ are the expected value of a sample group from such a probability distribution with uncertain variability ${\sigma }^2\left\{\text{a}100\right(1-$$\alpha )\mathrm{\%}$statistical power on $\mu$.

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

b) while $t\frac{\alpha }{2},n-1$is indeed the upper $100\frac{\alpha }{2}$ number of percent of the $t$probability with variables $n-1$,

$\frac{\alpha }{2}=\frac{1-0.95}{2}$

$=0.025$

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

$=1.99.$

To find the worth of $t$, we constructed a likelihood table for said Participant's $t$-distribution. A table depicts $t$-scores for variables (row) and confidence interval. The $t$-score is within the the chart where data matrix overlap.

The test statistic was $\overline{x}=6.425$, with a mean difference of $s=3.156$, as we expected.

$\begin{array}{r}6.425-1.99\frac{3.156}{\sqrt{83}}\le \mu \le 6.425+1.99\frac{3.156}{\sqrt{83}}\\ \end{array}$

$6.425-0.689\le \mu \le 6.425+0.689$

Graph ( c)

c)

• The$95\mathrm{\%}\mathrm{Cl}$for $\mu$is $5.736\le \mu \le 7.114$
• This problem's graph iswithin the following:

Error Bound (d)

d) - The cutoff point of such errors bound was

$EBM=0.689$

The precise mean analytics exam grade reflects in $95\%$ of all standard errors computed this approach. For instance , we could expected $95$of such confident areas to reflect its exact group mean if we created $100$of them.