## The National Assessment of Educational Progress (NAEP) includes a mathematics test for eighth-grade students. Scores on the test range from

The National Assessment of Educational Progress (NAEP) includes a mathematics test for eighth-grade students. Scores on the test range from 0 to 500. Demonstrating the ability to use the mean to solve a problem is an example of the skills and knowledge associated with performance at the Basic level. An example of the knowledge and skills associated with the Proficient level is being able to read and interpret a stem-and-leaf plot.In 2013, 170,100 8th-graders were in the NAEP sample for the mathematics test. The mean mathematics score was xbar = 285. We want to estimate the mean score μ in the population of all 8th-graders. Consider the NAEP sample as an SRS from a Normal population with standard deviation σ = 125.1) If we take many samples, the sample mean xbar 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?Please give your answer to 4 decimal places.2) According to the 95 part of the 68-95-99.7 rule, 95% of all values of x fall within ____ on either side of the unknown mean μ.What is the missing number?Give your answer to 4 decimal places3)The 95% confidence interval for the population mean score μ based on this one sample is between ____ and ____. (Give your answers to one decimal place.)

## Answers ( )

Answer:1) The standard deviation of the sampling distribution is:

√ V(X[bar])= 0.3033

2) 95% of the values on the normal distribution are between μ ± 0.6066

3) The formula for the confidence interval is:

[284.5; 285.5]

Step-by-step explanation:Hello!

Your study variable is

X: “score obtained on the test by an eighth-grader”

This variable has normal distribution:

X~N(μ;σ²)

Now the since the sample mean, X[bar], is obtained from n samples of the study variable, it is also a variable and has the same distribution as the original variable with the difference that is conditioned by the number of n samples from which it was calculated. So its distribution is:

X[bar]~N(μ;σ²/n)

Where E(X[bar])=μ and V(X[bar])= σ²/n

The sample information is:

n= 170100

X[bar]= 285

And the population standard deviation is known to be σ= 125.1

1) The standard deviation of the sampling distribution is:

√ V(X[bar]) = √(σ²/n) = σ/√n = 125.1/√170100 = 0.3033

2) 95% of the values on the normal distribution are between μ ± 2σ

This is μ ± 2*0.3033 = μ ± 0.6066

3) The formula for the confidence interval is:

X[bar] ± *σ/√n

285 ± 1.64 * 0.3033

[284.5; 285.5]

I hope it helps!