A company with two plants producing the same product, but with two slightly different manufacturing processes, conducted a sample study at each of the two plants with the following results: The sample statistics from the first group are: n = 40, x-bar = 18, s = 4.2. The sample statistics from the second group are: n = 50, x-bar = 20, s = 4. In testing the hypothesis that the mean time to produce the product is the same at both plants, what is the p-value of the test?
P-value = P(Z = -2.29) = 0.0219
To calculate the p-value for testing the hypothesis that the mean time to produce the product is the same at both plants, you need to perform a two-sample t-test.
Step 1: State the null and alternative hypotheses:
Null Hypothesis (H0): The mean time to produce the product is the same at both plants.
Alternative Hypothesis (HA): The mean time to produce the product is different at the two plants.
Step 2: Determine the significance level (alpha). Let's assume alpha = 0.05 (5%).
Step 3: Calculate the test statistic:
You can use the two-sample t-test formula:
t = (x1-bar - x2-bar) / sqrt((s1^2/n1) + (s2^2/n2))
Where:
x1-bar and x2-bar are the sample means,
s1 and s2 are the sample standard deviations,
n1 and n2 are the sample sizes.
In this case:
x1-bar = 18, s1 = 4.2, n1 = 40
x2-bar = 20, s2 = 4, n2 = 50
t = (18 - 20) / sqrt((4.2^2/40) + (4^2/50))
= -2 / sqrt((17.64/40) + (16/50))
= -2 / sqrt(0.44 + 0.32)
= -2 / sqrt(0.76)
= -2 / 0.8718
= -2.29 (rounded to two decimal places)
Step 4: Determine the degrees of freedom (df):
For a two-sample t-test, the degrees of freedom can be approximated using the formula:
df ≈ (s1^2/n1 + s2^2/n2)^2 / ( ((s1^2/n1)^2 / (n1-1)) + ((s2^2/n2)^2 / (n2-1)) )
Using the given values:
df ≈ ((4.2^2/40) + (4^2/50))^2 / ( ((4.2^2/40)^2 / (40-1)) + ((4^2/50)^2 / (50-1)) )
≈ (0.1764 + 0.128)^2 / ( ((0.1764^2/40) / 39) + ((0.16^2/50) / 49) )
≈ (0.3044)^2 / ( (0.00007728 / 39) + (0.0000512 / 49) )
≈ 0.09277 / (0.000001983 + 0.000001047 )
≈ 0.09277 / 0.00000303
≈ 30.64 (rounded to two decimal places)
Step 5: Calculate the p-value:
To calculate the p-value, we need to find the probability of observing a t-value as extreme as the calculated t-value (-2.29) or more extreme, assuming the null hypothesis is true (i.e., the mean time is the same at both plants).
Using a t-table or statistical software, you can find that the p-value for a two-tailed test with a t-value of -2.29 and degrees of freedom (df) of 30.64 is approximately 0.026.
Therefore, the p-value of the test is approximately 0.026.