. A company is considering installing new machines to assemble its products. The company is considering two types of machines, but it will buy only one type. The company selected eight assembly workers and asked them to use these two types of machines to assemble products. The following table gives the time taken (in minutes) to assemble one unit of the product on each type of machine for each of these eight workers.
Machine I 23 26 19 24 26 22 20 18
Machine II 21 24 23 25 24 25 24 23
Test at the 5% significance level if the mean time taken to assemble a unit of the product is different for the two types of machines.
Calculate the mean and standard deviation for each machine.
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√(n-1)
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to that Z score.
Hypothesis Test: Independent Groups (z-test)
Machine I Machine II
22.25 23.63 mean
3.06 1.30 std. dev.
8 8 n
-1.375 difference (Machine I - Machine II)
1.175 standard error of difference
0 hypothesized difference
-1.17 z
.2421 p-value (two-tailed)
To test if the mean time taken to assemble a unit of the product is different for the two types of machines, we can perform a paired t-test.
Here are the step-by-step instructions to perform the test:
Step 1: Define the null and alternative hypothesis:
- Null hypothesis (H0): The mean time taken to assemble a unit of the product is the same for both Machine I and Machine II.
- Alternative hypothesis (Ha): The mean time taken to assemble a unit of the product is different for Machine I and Machine II.
Step 2: Calculate the differences between the two types of machines for each worker. Subtract the time taken on Machine II from the time taken on Machine I.
Worker 1: 23 - 21 = 2
Worker 2: 26 - 24 = 2
Worker 3: 19 - 23 = -4
Worker 4: 24 - 25 = -1
Worker 5: 26 - 24 = 2
Worker 6: 22 - 25 = -3
Worker 7: 20 - 24 = -4
Worker 8: 18 - 23 = -5
Step 3: Calculate the mean and standard deviation of the differences.
Mean of the differences = (2 + 2 -4 -1 + 2 - 3 - 4 - 5) / 8 = -0.75
Standard deviation of the differences = √((2-(-0.75))^2 + (2-(-0.75))^2 + (-4-(-0.75))^2 + (-1-(-0.75))^2 + (2-(-0.75))^2 + (-3-(-0.75))^2 + (-4-(-0.75))^2 + (-5-(-0.75))^2) / 8 = 2.34
Step 4: Calculate the t-value.
t-value = (mean of the differences - hypothesized mean) / (standard deviation of the differences / √n)
We can assume the hypothesized mean is 0 (no difference).
t-value = (-0.75 - 0) / (2.34 / √8) = -0.75 / 0.829 = -0.905
Step 5: Determine the critical value and the p-value.
Since the question asks to test at the 5% significance level, we will use a two-tailed test. The critical value will be ±2.306 based on the t-distribution with 7 degrees of freedom.
The p-value can be calculated using the t-distribution table or statistical software. For this example, the p-value is approximately 0.397.
Step 6: Make a conclusion.
Comparing the t-value (-0.905) with the critical value (±2.306), we see that the t-value does not fall outside of the critical region. Additionally, the p-value of 0.397 is greater than the significance level of 0.05.
Therefore, we fail to reject the null hypothesis. We do not have enough evidence to conclude that the mean time taken to assemble a unit of the product is different for the two types of machines.
To test if the mean time taken to assemble a unit of the product is different for the two types of machines, we can use a paired t-test.
Here are the steps to perform the test:
Step 1: State the null hypothesis and alternative hypothesis:
- Null hypothesis (H0): There is no difference in the mean time taken to assemble a unit of the product between the two types of machines.
- Alternative hypothesis (Ha): There is a difference in the mean time taken to assemble a unit of the product between the two types of machines.
Step 2: Calculate the differences between the two machines for each worker:
- Subtract the time taken on Machine II from the time taken on Machine I for each worker.
Worker 1: 23 - 21 = 2
Worker 2: 26 - 24 = 2
Worker 3: 19 - 23 = -4
Worker 4: 24 - 25 = -1
Worker 5: 26 - 24 = 2
Worker 6: 22 - 25 = -3
Worker 7: 20 - 24 = -4
Worker 8: 18 - 23 = -5
Step 3: Calculate the mean and standard deviation of the differences:
- Mean (x̄) = (2 + 2 - 4 - 1 + 2 - 3 - 4 - 5) / 8 = -1
- Standard Deviation (s) = √( (2 - (-1))^2 + (2 - (-1))^2 + (-4 - (-1))^2 + (-1 - (-1))^2 + (2 - (-1))^2 + (-3 - (-1))^2 + (-4 - (-1))^2 + (-5 - (-1))^2 ) / (8 - 1) = √( 3^2 + 3^2 + (-3)^2 + 0^2 + 3^2 + (-2)^2 + (-3)^2 + (-4)^2 ) / 7 ≈ 3.08
Step 4: Calculate the t-value:
- t = (x̄ - 0) / (s / √n), where n is the number of workers (8)
t = (-1 - 0) / (3.08 / √8) ≈ -0.556
Step 5: Determine the critical t-value:
- Since the significance level is 5% (α = 0.05) and it is a two-tailed test, we need to find the critical t-value for a sample size of 8 and a significance level of 0.025 (0.05 / 2).
Using a t-table or a t-distribution calculator, the critical t-value is approximately ±2.36.
Step 6: Make a decision:
- If the calculated t-value falls within the rejection region (beyond the critical t-value), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
In this case, the calculated t-value (-0.556) does not fall beyond the critical t-value of ±2.36. Therefore, we fail to reject the null hypothesis.
Step 7: State the conclusion:
- Based on the test, there is not enough evidence to conclude that there is a difference in the mean time taken to assemble a unit of the product between the two types of machines at the 5% significance level.