91影视

It is given that 122 purchasing managers in the sample had a mean Mach rating score of 99.6, with a standard deviation of 12.6. i.e.$n=122,\stackrel{炉}{x}=99.6,蟽=12.6$

a.

Since, sample mean is unbiased estimator of population mean.

So, the true mean Mach rating score of all purchasing managers is

$\begin{array}{rcl}渭& =& \stackrel{炉}{x}\\ & =& 99.6\\ & & \end{array}$

b.

The 95% confidence interval for population mean is given by

$\left(\stackrel{炉}{x}-E<渭<\stackrel{炉}{x}+E\right)$

The above figure shows that the area in each of the red-shaded tails is$\frac{伪}{2}=0.025$

Now, find $z_{\frac{伪}{2}}=1.96$by noting that the cumulative area to its left must be 1-0.025 or 0.975.

Now, refer to the standard normal table values to find the area of 0.975 corresponds to$z=1.96$

For the 95% confidence level, the critical value is therefore$z_{\frac{伪}{2}}=1.96$

To find the critical z score for the 95% confidence level, look up 0.975 in the body of the table.

Now, the margin of error is calculated by using the following formula:

$E=z_{\frac{伪}{2}}脳\frac{蟽}{\sqrt{n}}$with $z_{\frac{伪}{2}}=1.96$

$\begin{array}{rcl}E& =& 1.96脳\frac{12.6}{\sqrt{122}}\\ & =& 1.96脳1.1408\\ & =& 2.236\\ & & \end{array}$

Simply substitute and E in the general format for confidence interval is as follows:

$$\begin{gathered} \stackrel{炉}{x}-E<渭<\stackrel{炉}{x}+E \\ 99.6-2.2359<渭<99.6+2.2359 \\ 97.364<渭<101.839 \end{gathered}$$

c.

It can conclude that it is 95% confident that the true mean Mach rating score of all purchasing managers will lie between 97.364 and 101.839.

d.

Given that a director of purchasing at a major firm claims that the true mean Mach rating score of all purchasing managers is 85.

The evidence that the true mean Mach rating score of all purchasing managers is 85.

Because Mach rating score belong to the 95% confidence interval for the true mean Mach rating score of all purchasing managers.