91影视

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 $\stackrel{炉}{X}$is the mean number of costs from a sample of $84$ used car sales. If $\stackrel{炉}{x}$and $\mathrm{S}$are the mean and standard deviation of a random sample from a normal distribution with unknown variance ${蟽}^2,\$\backslash \text{textbf}\left\{\text{a}100\right(1-伪)\%$confidence interval on$渭\}$ is given by

$\stackrel{炉}{x}-t_{\frac{伪}{2},n-1}\frac{s}{\sqrt{n}}鈮�渭鈮�\stackrel{炉}{x}+t_{\frac{伪}{2},n-1}\frac{s}{\sqrt{n}}$

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

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

$\frac{伪}{2}=\frac{1-0.95}{2}=0.025$$t_{\frac{伪}{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

• A $95\%$CI for$渭$is

$5.736鈮�渭鈮�7.114$

• The graph of this problem is

• 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.