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a_sh-v [17]
3 years ago
10

John Calipari, head basketball coach for the 2012 national champion University of Kentucky Wildcats, is the highest paid coach i

n college basketball with an annual salary of $5.4 million (USA Today, March 29, 2012). The sample below shows the head basketball coaches salary for a sample of 10 schools playing NCAA Division 1 basketball. Salary data are in millions of dollars. University Coach’s Salary University Coach’s Salary Indiana 2.2 Syracuse 1.5 Xavier .5 Murray State .2 Texas 2.4 Florida State 1.5 Connecticut 2.7 South Dakota State .1 West Virginia 2.0 Vermont .2 a. Use the sample mean for the 10 schools to estimate the population mean annual salary for head basketball coaches at colleges and universities playing NCAA Division 1 basketball (to 2 decimals). b. Use the data to estimate the population standard deviation for the annual salary for head basketball coaches (to 4 decimals). c. What is the 95% confidence interval for the population variance (to 2 decimals)
Mathematics
1 answer:
MrRissso [65]3 years ago
7 0

Answer:

A) sample mean = $1.36 million

B) standard deviation = $0.9189 million

C) confidence interval = ($1.93 million , $0.79 million)

*since the sample size is very small, the confidence interval is not valid.

Step-by-step explanation:

samples:

  1. $2.7 million
  2. $2.4 million
  3. $2.2 million
  4. $2 million
  5. $1.5 million
  6. $1.5 million
  7. $0.5 million
  8. $0.5 million
  9. $0.2 million
  10. $0.1 million

sample mean = $1.36 million

the standard deviation:

  • $2.7 million - $1.36 million = 1.34² = 1.7956
  • $2.4 million - $1.36 million = 1.04² = 1.0816
  • $2.2 million - $1.36 million = 0.84² = 0.7056
  • $2 million - $1.36 million = 0.64² = 0.4096
  • $1.5 million - $1.36 million = 0.14² = 0.0196
  • $1.5 million - $1.36 million = 0.14² = 0.0196
  • $0.5 million - $1.36 million = -0.86² = 0.7396
  • $0.5 million - $1.36 million = -0.86² = 0.7396
  • $0.2 million - $1.36 million = -1.16² = 1.3456
  • $0.1 million - $1.36 million = -1.26² = 1.5876
  • total $8.444 million / 10 = $0.8444 million

standard deviation = √0.8444 = 0.9189

95% confidence interval = mean +/- 1.96 standard deviations/√n:

$1.36 million + [(1.96 x $0.9189 million)/√10] = $1.36 million + $0.57 million = $1.93 million

$1.36 million - $0.57 million = $0.79 million

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Answer:

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Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

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The sampling distribution of the sample proportion of adults who have credit card debts of more than $2000 is approximately normally distributed with mean \mu = 0.39 and standard deviation s = 0.0488

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