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!

3, 12, 48, 384, 1536, 6144, 24576, 98304

step-by-step explanation:

just keep timing it by 4.