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)
Answer:
A) x¯ = $1.33 million
B) σ = $0.9508 million
C) CI = ($0.74 million, $1.92 million)
Explanation:
We are given the salary data as;
$2.2 million, $1.5 million, $0.5 million, $0.2 million, $2.4 million, $1.5 million, $2.7 million, $0.1 million, $2 million, $0.2 million
A) Population mean annual salary = $(2.2 + 1.5 + 0.5 + 0.2 + 2.4 + 1.5 + 2.7 + 0.1 + 2 + 0.2)/10 in millions
x¯ = $1.33 million
B) Formula for the standard deviation is;
√(Σ(x - x¯)²/n)
Thus, let's first find (x - x¯)² all in millions
>> (2.7 - 1.33)² = 1.37² = 1.8769
>> (2.4 - 1.33)² = 1.07² = 1.1449
>> (2.2 - 1.33)² = 0.87² = 0.7569
>> (2 - 1.33)² = 0.67² = 0.4489
>> (1.5 - 1.33)² = 0.17² = 0.0289
>> (1.5 - 1.33)² = 0.17² = 0.0289
>> (0.5 - 1.33)² = (-0.83)² = 0.6889
>> (0.2 - 1.33)² = (-1.13)² = 1.2769
>> (0.2 - 1.33)² = (-1.13)² = 1.2769
>> (0.1 - 1.33)² = (-1.23)² = 1.5129
Σ(x - x¯)² = 1.8769 + 1.1449 + 0.7569 + 0.4489 + 0.0289 + 0.0289 + 0.6889 + 1.2769 + 1.2769 + 1.5129
Σ(x - x¯)² = $9.041 million
Σ(x - x¯)²/n = $9.041/10 = $0.9041 million
standard deviation = √0.9041
σ = $0.9508 million
C) Critical value at a Confidence level of 95% is z = 1.96
Thus,formula for critical interval is;
CI = x¯ ± z(σ/√n)
CI = 1.33 ± 1.96(0.9508/√10)
CI = ($0.74 million, $1.92 million)