Employees in a large computer firm claim that the mean salary of the firm’s programmers is less than that of its competitors. The competitor’s salary is $47,000. A random sample of 30 of the firm’s programmers has a mean salary of $46,500 with a standard deviation of 5500. Calculate the test statistic for the hypothesis: Ho: mean >= 47000, H1: mean < 47000
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√(n-1)
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to that Z score.
To calculate the test statistic for the hypothesis, we need to perform a one-sample t-test. The formula for the test statistic is:
t = (x̄ - μ) / (s / √n)
Where:
x̄ = sample mean
μ = population mean (hypothesized mean)
s = sample standard deviation
n = sample size
In this case:
x̄ = $46,500
μ = $47,000
s = $5,500
n = 30
Now, let's substitute the values into the formula and calculate the test statistic:
t = (46,500 - 47,000) / (5,500 / √30)
t = (-500) / (5,500 / √30)
To compute the value of the denominator, we need to find the square root of 30 first:
√30 = 5.48 (rounded to two decimal places)
Now, we can calculate the test statistic:
t = (-500) / (5,500 / 5.48)
t = (-500) / (1,600.4)
t ≈ -0.312 (rounded to three decimal places)
Therefore, the test statistic for the given hypothesis is approximately -0.312.