Statistics for Business & Economics, Revised (MindTap Course List)
Statistics for Business & Economics, Revised (MindTap Course List)
12th Edition
ISBN: 9781285846323
Author: David R. Anderson, Dennis J. Sweeney, Thomas A. Williams, Jeffrey D. Camm, James J. Cochran
Publisher: South-Western College Pub
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Textbook Question
Chapter 7.5, Problem 26E

The mean annual cost of automobile insurance is $939 (CNBC, February 23, 2006). Assume that the standard deviation is σ = $245.

  1. a. What is the probability that a sample of automobile insurance policies will have a sample mean within $25 of the population mean for each of the following sample sizes: 30, 50, 100, and 400?
  2. b. What is the advantage of a larger sample size when attempting to estimate the population mean?

a.

Expert Solution
Check Mark
To determine

Find the probability that a sample of automobile insurance policies will have a sample mean within $25 of the population mean for each of the following sample sizes: 30, 50, 100, and 400.

Answer to Problem 26E

The probability that a sample of automobile insurance policies will have a sample mean within $25 of the population mean for the sample size 30 is 0.4246.

The probability that a sample of automobile insurance policies will have a sample mean within $25 of the population mean for the sample size 50 is 0.5284.

The probability that a sample of automobile insurance policies will have a sample mean within $25 of the population mean for the sample size 100 is 0.6922.

The probability that a sample of automobile insurance policies will have a sample mean within $25 of the population mean for the sample size 400 is 0.9586.

Explanation of Solution

Calculation:

The standard deviation is σ=$245 and the mean annual cost of automobile insurance is $939.

Sampling distribution of x¯:

  • The expected value of x¯ is, E(x¯)=μ.
  • The standard deviation of x¯ is

    For finite population, σx¯=NnN1(σn)

    For infinite population, σx¯=σn

  • When the population size is infinite or finite and nN0.05 then the standard deviation of x¯ is σx¯=σn.

The expected value of x¯ is obtained below:

E(x¯)=μ=$939

Thus, the expected value of x¯ is $939.

For sample size 30:

The standard deviation is σx¯=σn

Substitute σ as $245 and n as 30 in the formula,

σx¯=σn=24530=2455.4772=44.73

Thus, the standard deviation of x¯ is $44.73.

Therefore, the sampling distribution of the sample mean x¯ is approximately normal with μ=$939 and σx¯=$44.73.

The probability that a sample of automobile insurance policies will have a sample mean within $25 of the population mean for the sample size 30 is calculated as follows:

P(|x¯|μ±25)=P(μ25x¯μ+25)=P[25σx¯x¯μσx¯25σx¯]=P(2544.73z2544.73)=P(0.56z0.56)=P(z0.56)P(z0.56)

For z=0.56:

From Table 1: Cumulative probabilities for the standard normal distribution,

  • Locate the value 0.5 in the first column.
  • Locate the value 0.06 in the first row corresponding to the value 0.5 in the first column.
  • The intersecting value of row and column is 0.7123.

For z=0.56:

From Table 1: Cumulative probabilities for the standard normal distribution,

  • Locate the value –0.5 in the first column.
  • Locate the value 0.06 in the first row corresponding to the value –0.5 in the first column.
  • The intersecting value of row and column is 0.2877.

Therefore,

P(μ25x¯μ+25)=0.71230.2877=0.4246

Thus, the probability that a sample of automobile insurance policies will have a sample mean within $25 of the population mean for the sample size 30 is 0.4246.

For sample size 50:

The standard deviation is obtained below:

Substitute σ as $245 and n as 50 in the formula.

σx¯=σn=24550=2457.0711=34.65

Thus, the standard deviation of x¯ is 34.65.

Therefore, the sampling distribution of the sample mean x¯ is approximately normal with μ=$939 and σx¯=$34.65.

The probability that a sample of automobile insurance policies will have a sample mean within $25 of the population mean for the sample size 50 is calculated as follows:

P(|x¯|μ±25)=P(μ25x¯μ+25)=P[25σx¯x¯μσx¯25σx¯]=P(2534.65z2534.65)=P(0.72z0.72)=P(z0.72)P(z0.72)

For z=0.72:

From Table 1: Cumulative probabilities for the standard normal distribution,

  • Locate the value 0.7 in the first column.
  • Locate the value 0.02 in the first row corresponding to the value 0.7 in the first column.
  • The intersecting value of row and column is 0.7642.

For z=0.72:

From Table 1: Cumulative probabilities for the standard normal distribution,

  • Locate the value – 0.7 in the first column.
  • Locate the value 0.02 in the first row corresponding to the value –0.7 in the first column.
  • The intersecting value of row and column is 0.2358.

Therefore,

P(μ25x¯μ+25)=0.76420.2358=0.5284

Thus, the probability that a sample of automobile insurance policies will have a sample mean within $25 of the population mean for the sample size 50 is 0.5284.

For sample size 100:

The standard deviation is obtained below:

Substitute σ as $245 and n as 100 in the formula.

σx¯=σn=245100=24510=24.5

Thus, the standard deviation of x¯ is $24.5.

The sampling distribution of the sample mean x¯ is approximately normal with μ=$939 and σx¯=$24.5.

The probability that a sample of automobile insurance policies will have a sample mean within $25 of the population mean for the sample size 100 is calculated as follows:

P(|x¯|μ±25)=P(μ25x¯μ+25)=P[25σx¯x¯μσx¯25σx¯]=P(2524.5z2524.5)=P(1.02z1.02)=P(z1.02)P(z1.02)

For z=1.02:

From Table 1: Cumulative probabilities for the standard normal distribution,

  • Locate the value 1.0 in the first column.
  • Locate the value 0.02 in the first row corresponding to the value 1.0 in the first column.
  • The intersecting value of row and column is 0.8461.

For z=1.02:

From Table 1: Cumulative probabilities for the standard normal distribution,

  • Locate the value –1.0 in the first column.
  • Locate the value 0.02 in the first row corresponding to the value –1.0 in the first column.
  • The intersecting value of row and column is 0.1539.

Therefore,

P(μ25x¯μ+25)=0.84610.1539=0.6922

Thus, the probability that a sample of automobile insurance policies will have a sample mean within $25 of the population mean for the sample size 100 is 0.6922.

For sample size 400:

The standard deviation is obtained below:

Substitute σ as $245 and n as 400 in the formula.

σx¯=σn=245400=24520=12.25

Thus, the standard deviation of x¯ is $12.25.

The sampling distribution of the sample mean x¯ is approximately normal with μ=$939 and σx¯=$12.25.

The probability that a sample of automobile insurance policies will have a sample mean within $25 of the population mean for the sample size 400 is calculated as follows:

P(|x¯|μ±25)=P(μ25x¯μ+25)=P[25σx¯x¯μσx¯25σx¯]=P(2512.25z2512.25)=P(2.04z2.04)=P(z2.04)P(z2.04)

For z=2.04:

From Table 1: Cumulative probabilities for the standard normal distribution,

  • Locate the value 2.0 in the first column.
  • Locate the value 0.04 in the first row corresponding to the value 2.0 in the first column.
  • The intersecting value of row and column is 0.9793.

For z=2.04:

From Table 1: Cumulative probabilities for the standard normal distribution,

  • Locate the value –2.0 in the first column.
  • Locate the value 0.04 in the first row corresponding to the value –2.0 in the first column.
  • The intersecting value of row and column is 0.0207.

Therefore,

P(μ25x¯μ+25)=0.97930.0207=0.9586

Thus, the probability that a sample of automobile insurance policies will have a sample mean within $25 of the population mean for the sample size 400 is 0.9586.

b.

Expert Solution
Check Mark
To determine

Explain the advantage of a larger sample size to estimate the population mean.

Explanation of Solution

From part a, it was observed that the probability of being with ±25 of μ is from 0.4246 for sample size 30 to 0.9586 for sample size 400.

Hence, it can be concluded that as the sample size increases, the probability becomes closer to 1.

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Chapter 7 Solutions

Statistics for Business & Economics, Revised (MindTap Course List)

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