A school district claims that the average teacher in the district earns $45,000 per year. The teacher's union disputes this claim and argues that the average salary is actually less. A random sample of 20 teachers yields a mean salary of $44,500 with a sample standard deviation of $1,750. What's the P-value for
a test of the hypothesis that H_{0} : mu = 45 and H₂: μ < 45,000?
A. 0.025 < P < 0.05
B. 2 < P < 0.025
5 < P < 0.1
D. 0.01 < P < 0.02
10 < P < 15
To find the p-value for the hypothesis test, we can use the t-test formula:
t = (x̄ - μ) / (s / √n)
Where:
x̄ = sample mean ($44,500)
μ = population mean ($45,000)
s = sample standard deviation ($1,750)
n = sample size (20)
t = (44,500 - 45,000) / (1,750 / √20)
t = -500 / (1,750 / 4.472)
t = -1.143
To find the p-value, we can use a t-distribution table or a calculator. Since we are testing for μ < 45,000, this is a one-tailed test.
With a t-statistic of -1.143 and 19 degrees of freedom (n-1 = 20-1 = 19), the p-value is approximately 0.133.
Therefore, the correct answer is not listed in the options provided. The p-value for the test of the hypothesis H₀: μ = 45 and H₂: μ < 45,000 is approximately 0.133.