An issue of Scientific American revealed that the batting averages of major-league baseball players are normally distributed with mean \(.270\) and standard deviation \(.031\).

a. Simulate \(2000\) samples of five batting averages each.

b. Determine the sample mean and sample standard deviation of each of the \(2000\) samples.

c. For each of the \(2000\) samples, determine the observed value of the standardized version of \(\bar{x}\).

d. Obtain a histogram of the \(2000\) observations in part (c).

e. Theoretically, what is the distribution of the standard version of \(\bar{x}\)?

f. Compare your results from parts (d) and (e).

g. For each of the \(2000\) samples, determine the observed value of the studentized version of \(\bar{x})\).

h. Obtain a histogram of the \(2000\) observations in part (g).

i. Theoretically, what is the distribution of the standard version of \(\bar{x})\) ?

j. Compare your results from parts (h) and (i).

k. Compare your histograms from parts (d) and (h). How and why do they differ?

Short Answer

Expert verified

Part a. The 2000 random sample is

Then the five batting average will be

Part b. \(\bar{x}=0.2698\)

\(\sigma=0.0134\)

Part c.

Part d.

Step by step solution

01

Part a. Step 1. Given information

The number of sample \((n)\), mean \((\bar{x})\) and standard deviation \((\sigma)\) is given.

\(\bar{x}=0.27\)

\(\sigma=0.031\)

\(n=2000\times 5=10000\)

02

Part a. Step 2. Calculation 

Generate \(2000\) samples of males using function “norminv” with \(0.031\) sample mean and \(0.27\) standard deviation in MATLAB

\(r=norminv(rand(2000,5), 0.27, 0.031)\)

After that we will get random sample \(2000\) males.

Program:

Query:

  • First, we have defined the number of samples.
  • Then create \(2000\) random samples using function “norminv” with sample mean \(0.27\) and standard deviation \(0.031\).
03

Part b. Step 1. Calculation

Calculate the mean using relation

\(\bar{x}=\frac{\sum_{i-1}^{n}x_{i}}{n}\)

After simplifying, we will get

\(\bar{x}=0.2698\)

Calculate the standard deviation using relation

\(\sigma =\sqrt{\frac{\sum (x_{i}-\mu)^{2}}{n}}\)

After simplifying we will get

\(\sigma=0.0134\)

The solution is \(\bar{x}=0.2698\)

\(\sigma=0.0134\)

04

Part c. Step 1. Calculation

Calculate the mean using relation

\(\bar{x}=\frac{\sum_{i-1}^{n}x_{i}}{n}\)

After simplifying, we will get

\(\bar{x}=0.2698\)

Calculate the standard deviation using relation

\(\sigma =\sqrt{\frac{\sum (x_{i}-\mu)^{2}}{n}}\)

After simplifying we will get

\(\sigma=0.0134\)

Calculate the standardized value for the all samples

\(z=\frac{sample mean-0.2698}{0.0134}\)

Program:

Query:

  • First, we have defined the number of samples.
  • Then generate the random samples.
  • Then calculate the mean and standard deviation.
  • Calculate the z-score which is standardized form of all samples.
05

Part d. Step 1. Calculation

Calculate the mean using relation

\(\bar{x}=\frac{\sum_{i-1}^{n}x_{i}}{n}\)

After simplifying, we will get

\(\bar{x}=0.2698\)

Calculate the standard deviation using relation

\(\sigma =\sqrt{\frac{\sum (x_{i}-\mu)^{2}}{n}}\)

After simplifying we will get

\(\sigma=0.0134\)

Calculate the standardized value for the all samples

\(z=\frac{sample mean-0.2698}{0.0134}\)

Program:

Query:

  • First, we have defined the number of samples.
  • Then generate the random samples.
  • Then calculate the mean and standard deviation.
  • Calculate the z-score which is standardized form of all samples.
  • Then generate the histogram.

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Most popular questions from this chapter

M\&Ms. In the article "Sweetening Statistics-What M\&M's Can Teach Us" (Minitab Inc., August 2008), M. Paret and E. Martz discussed several statistical analyses that they performed on bags of M\&Ms. The authors took a random sample of 30 small bags of peanut M\&Ms and obtained the following weights, in grams (g).

a. Determine a 95%lower confidence bound for the mean weight of all small bags of peanut M\&Ms. (Note: The sample mean and sample standard deviation of the data are 52.040gand 2.807grespectively.)

b. Interpret your result in pant (a).

c. According to the package, each small bag of peanut M\&Ms should weigh 49.3gComment on this specification in view of your answer to part (b) It provides equal confidence with a greater lower limit.

Part (c) Because the weight of 49.3g is below the 95% lower confidence bound.

American Alligators. Refer to Exercise 8.78.

a. Determine the margin of error for the 95%confidence interval.

b. Determine the margin of error for the 99%confidence interval.

c. Compare the margins of error found in parts (a) and (b).

d. What principle is being illustrated?

Assume that the population standard deviation is known and decide weather use of the z-interval procedure to obtain a confidence interval for the population mean is reasonable. Explain your answers.

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a. t0 as

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When estimating an unknown parameter, what does the margin of error indicate?

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