Q. 5 - The Practice of Statistics for AP

Question

5. NAEP scores Young people have a better chance of full-time employment and good wages if they are good with numbers. How strong are the quantitative skills of young Americans of working age? One source
of data is the National Assessment of Educational Progress (NAEP) Young Adult Literacy Assessment Survey, which is based on a nationwide probability sample of households. The NAEP survey includes a
short test of quantitative skills, covering mainly basic arithmetic and the ability to apply it to realistic problems. Scores on the test range from $0$ to $500$. For example, a person who scores $233$ can add the amounts of two checks appearing on a bank deposit slip; someone scoring $325$ can determine the price of a meal from a menu; a person scoring $375$ can
transform a price in cents per ounce into dollars per pound. Suppose that you give the NAEP test to an SRS of $840$ people from a large population in which the scores have mean $280$ and standard deviation$\sigma =60$. The mean $\overline{x}$ of the $840$ scores will vary if you take repeated samples.
(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 population mean $\mu$ lies in the confidence interval$\overline{x}\pm m$. For what percent of all possible samples does the interval capture $\mu$?

Step-by-Step Solution

Answer

(a) Approximately normal with mean $280$ as center and the standard deviation is $2.0702$.

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

(c) The distance $m$ of the mean of the sampling distribution is $4.140$.

(d) All possible samples $\mu$captures $95\%$.

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 sample mean as  $\overline{x}=280$.

The standard deviation as $\sigma =60$
And the Sample size as $n=840$
Use the Central Limit to determine the shape of the sampling distribution.
As a result, the sampling distribution of the sample mean has a normal shape.
The mean of the sampling distribution of the sample mean $\mu =280$ is calculated using the information provided.
So, the center equal to the mean $\mu =280$
And for the spread, use the formula
Population standard deviation $=\frac{\text{ Standard deviation }}{\text{ Square root of sample size }}$

${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$

$=\frac{60}{\sqrt{840}}$

$\approx 2.0702$

As a result, the shape is approximately normal with mean $280$as center and the spread is the standard deviation which is equal $2.0702$.

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 Sample mean of the score$\overline{x}=280$
The standard deviation $\sigma =60$
And the sample size of the people from the large population $n=840$.

Part (b) Step 3: Explanation

The sample mean $\overline{x}$ is then normally distributed with mean $\mu =280$ and standard deviation is calculated as:
$\frac{\sigma }{\sqrt{n}}=\frac{60}{\sqrt{840}}$

$\approx 2.070$.

Part (c) Step 1: Given information

To determine the distance$m$ of the mean of the sampling distribution.
Then shade the region on the axis of the sketch that is within $m$ of the mean.

Part (c) Step 2: Explanation

Let, the sample mean is $\overline{x}=280$.

The standard deviation is $\sigma =60$.
And the Sample size is $n=840$
Determine the standard deviation as follows:
${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$
$=\frac{60}{\sqrt{840}}$
$\approx 2.0702$
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.
Hence,

$m=2\sigma$
$=2\times 2.070$
$=4.140$

Part (c) Step 3: Explanation

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

As a result, the distance $m$ of the mean of the sampling distribution is $4.140$.

Part (d) Step 1: Given information

The population mean $\mu$ lies in the confidence interval $\overline{x}\pm m$. To determine the percent of all possible samples $\mu$captured by the interval.

Part (d) Step 2: Explanation

Since, the sample mean of the score is $\overline{x}=280$.
The standard deviation is $\sigma =60$.
And the sample size of the people from the large population is $n=840$.
The mean of the sampling distribution is distance $m$.
Approximately $95$ percent  of all $\overline{x}$ values are within $m$of the sampling distribution's mean.
As a result, $\mu$ captures $95\%$ of all possible samples.