Based on a random sample of 25 units of lawnmowers, the average weight is 102 lbs., and the sample standard deviation is 10 lbs. We would like to decide if there is enough evidence to establish that the average weight for the population of lawnmowers is greater than 100 lbs. Assume the population is normally distributed. What is the sample value of the test statistic?
Try a one-sample t-test (your sample size is fairly small).
To determine the sample value of the test statistic, we need to calculate the z-score. The z-score tells us how many standard deviations the sample mean is away from the population mean.
The formula for the z-score is:
z = (X̄ - μ) / (σ / √n)
Where:
X̄ = sample mean (102 lbs.)
μ = population mean (100 lbs.)
σ = standard deviation of the population (unknown)
n = sample size (25 units)
Since we are given the sample standard deviation (10 lbs.), we can use it as an estimate for the population standard deviation (σ).
Plugging in the given values into the formula, we have:
z = (102 - 100) / (10 / √25)
= 2 / (10 / 5)
= 2 * 5 / 10
= 1
Therefore, the sample value of the test statistic is 1.