The National Assessment of Educational Progress (NAEP) gave a test of basic arithmetic and the ability to apply it in everyday life to a sample of 840 men 21 to 25 years of age. Scores range from 0 to 500; for example, someone with a score of 325 can determine the price of a meal from a menu. The mean score for these 840 young men was x¯¯¯ = 272. We want to estimate the mean score μ in the population of all young men. Consider the NAEP sample as an SRS from a Normal population with standard deviation σ = 60. (a) If we take many samples, the sample mean x¯¯¯ varies from sample to sample according to a Normal distribution with mean equal to the unknown mean score μ in the population. What is the standard deviation of this sampling distribution? (b) According to the 68 part of the 68-95-99.7 rule, 68% of all values of x¯¯¯ fall within on either side of the unknown mean μ. What is the missing number? (c) What is the 68% confidence interval for the population mean score μ based on this one sample? Note: Use the 68-95-99.7 rule to find the interval.

Step-by-step explanation:

Hello!

There were a basic arithmetic test and the ability to apply it in every life given to 840 men (21 - 25 years old), the scores range is from 0 to 500.

The means score of the sample was X[bar]= 272

The study variable is X: Score on a basic arithmetic test obtained by a man 21-25 years old.

This variable has a normal distribution with known population standard deviation of σ= 60

X~N(μ;σ²)

a.

If X₁, X₂, ..., Xₙ be the n random variables that constitute a sample, then any function of type θ = î (X₁, X₂, ..., Xₙ) that depends solely on the n variables and does not contain any parameters known, it is called the estimator of the parameter.

When the function i (.) It is applied to the set of the n numerical values of the respective random variables, a numerical value is generated, called parameter estimate θ.

This follows the concepts:

1) The function i (.) It is a function of random variables, so it is also a random variable, that is to say, that every estimator is a random variable.

2) From the above, it follows that Î has its probability distribution and therefore mathematical hope, E (î), and variance, V (î).

This means that if you take many samples of the same population and calculate their sample mean, the sample mean X[bar] will be a random variable that shares the same distribution qualities as the original population.

So our variable of interest has a normal distribution X~N(μ;σ²)

From this, we can say that X[bar]~N(μ;σ²/n)

The variance of the sample mean is σ²/n ⇒ the standard deviation is its square root σ/√n

Numerically: σ/√n= 60/√840= 2.07

b.

The empirical rule states that:

68% of the data under a normal distribution lies within μ ± δ

95% of the data under a normal distribution lies within μ ± 2δ

99% of the data under a normal distribution lies within μ ± δ.

So under the distribution of the sample mean you'd expect that 68% of all values will be within ± 2.07

c.

To calculate this 68% CI using the empirical rule you have that

[X[bar] ± σ/√n]

X[bar]= 272

σ/√n= 2.07

[272 ± 2.07]

68% CI is [269.93; 274.07]

I hope it helps!

when rotating a figure 180 degrees, imagine that you are able to take the figure and turn it clockwise or counterclockwise around a center point. to rotate a figure 180 degrees, you will need to apply the rule (x, y) → (-x, -y). start by using a coordinate grid with coordinates for each vertex of the figure. the center point of the coordinate grid is located at (0, 0), which is what you will rotate the figure around. write down the original coordinates of the shape you are going to rotate. then, apply the rule. for example, coordinate (1, 2) becomes (-1, -2). another example would be, coordinate (-4, -2) becoming (4, 2). after you change each original coordinate to the rotated coordinate, you will draw your new figure.

step-by-step explanation:

step-by-step explanation:

graph using a few selected

x         y

-1       15

0         6

1         2.4

2       0.96