Question

6. Auto emissions Oxides of nitrogen (called $NOX$for short) emitted by cars and trucks are important contributors to air pollution. The amount of $NOX$
emitted by a particular model varies from vehicle to vehicle. For one light-truck model, $NOX$emissions vary with mean $\mu$that is unknown and standard deviation $\sigma =0.4$gram per mile. You test an $SRS$of $50$of these trucks. The sample mean $NOX$level $\overline{x}$estimates the unknown $\mu$. You will get different values of $\overline{x}$if you repeat your sampling.

(a) Describe the shape, center, and spread of the sampling distribution of $\overline{x}$.
(b) Sketch the sampling distribution of $\overline{x}$. Mark its mean and the values one, two, and three standard deviations on either side of the mean.
(c) According to the $68–95–99.7$rule, about $95\%$of all values of $\overline{x}$lie within a distance $m$of the mean of the sampling distribution. What is $m$? Shade the region on the axis of your sketch that is within $m$ of the mean.
(d) Whenever $\overline{x}$falls in the region you shaded, the unknown population mean $\mu$lies in the confidence interval $\overline{x}\pm m$. For what percent of all possible samples does the interval capture $\mu$?

Short answer

(a) The shape is approximately normally distributed with center$\mu$and spread is $0.057$.

(b) Sketched the sampling distribution of $\overline{x}$as:

(c) The distance $m$of the mean of the sampling distribution is $0.114$.

(d) The $95\%$of all possible sample capture $\mu$.

Step-by-step solution

Part (a) Step 1: Given information

To describe the shape, center, and spread of the sampling distribution of $\overline{x}$.

Part (a) Step 2: Explanation

Let, the number of trucks tested for $NOX$emission test is $n=50$.

And the standard deviation is $\sigma =0.04$.

Use the Central Limit Theorem to determine the shape of the sample distribution $\left(CLT\right)$. The sample mean $\overline{x}$is nearly normally distributed in terms of sampling distribution. The sample mean for the center is $\mu$, based on the sampling distribution information provided.
Population standard deviation $=\frac{\text{Standarddeviation}}{\text{Square rootof samplesize}}$
${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$
$=\frac{0.4}{\sqrt{50}}$
$\approx 0.057$
As a result, the shape is approximately normally distributed with center $\mu$and spread is $0.057$.

Part(b) Step 1: Given information

To sketch the sampling distribution of $\overline{x}$. Then to mark its mean and the values one, two, and three standard deviations on either side of the mean.

Part (b) Step 2: Explanation

Let, the number of trucks tested for $NOX$emission test is $n=50$.

And the standard deviation $\sigma =0.04$.

Part (b) Step 3: Explanation

The sample mean $\overline{x}$is normally distributed with the mean $\mu$ and the standard deviation as:
$\frac{\sigma }{\sqrt{n}}=\frac{0.4}{\sqrt{50}}$

$\approx 0.057$

Part (c) Step 1: Given information

To determine the distance $m$. And to shade the region on the axis of the sketch that is within $m$ of the mean.

Part (c) Step 2: Explanation

Let, the number of trucks tested for$NOX$emission test is $n=50$.

The standard deviation is $\sigma =0.04$

And the population standard deviation ${\sigma }_x=0.057$.

According to the $68-95-00.7$percent rule, around $95$percent of all $\overline{x}$values are within $m$ of the mean, or twice the population standard deviations of the mean.
$m=2\sigma$
$=2\times 0.057$
$=0.114$

Part (c) Step 3: Explanation

Shade the region on the axis of the sketch that is within $m$of the mean as:

As a result, distance $m$of the mean of the sampling distribution is $0.114$.

Part (d) Step 1: Given information

To determine the percent of all possible samples does the interval capture $\mu$.

Part (d) Step 2: Explanation

Let, the number of trucks tested for $NOX$emission test is $n=50$.
And the standard deviation $\sigma =0.04$.
Since, $95$percent of the $\overline{x}$values from part c are within a distance $m$ of the sample distribution's mean.
As a result, $95\%$of all possible sample capture $\mu$.