1. The average number of articles produced by two machines per day is 200 and 250 with standard deviation 20 and 25 respectively on the basis of records of 25 days production. Can you regard both the machines equally efficient at 1% level of significance?
To determine whether both machines are equally efficient, we can perform a hypothesis test. The null hypothesis (H0) is that the machines are equally efficient, while the alternative hypothesis (Ha) is that the machines are not equally efficient.
To conduct the hypothesis test, we will perform a two-sample t-test. However, before performing the test, we need to calculate the test statistic, which measures the difference between the sample means relative to the variability within the samples.
Here are the steps to calculate the test statistic:
Step 1: Define the null and alternative hypotheses:
- Null hypothesis (H0): The machines are equally efficient (μ1 = μ2).
- Alternative hypothesis (Ha): The machines are not equally efficient (μ1 ≠ μ2).
Step 2: Select the significance level (α):
In this case, the significance level is given as 1% or 0.01.
Step 3: Collect the data:
We are given the following information:
- Number of days (n): 25
- Machine 1:
- Sample mean (x̄1): 200
- Standard deviation (s1): 20
- Machine 2:
- Sample mean (x̄2): 250
- Standard deviation (s2): 25
Step 4: Calculate the test statistic:
We will use the two-sample t-test formula to calculate the test statistic:
t = (x̄1 - x̄2) / sqrt((s1²/n1) + (s2²/n2))
Where:
- x̄1 and x̄2 are the sample means
- s1 and s2 are the standard deviations
- n1 and n2 are the sample sizes
Plugging in the values, we get:
t = (200 - 250) / sqrt((20²/25) + (25²/25))
Simplifying this expression gives us:
t = -50 / sqrt((16/25) + 25)
Step 5: Determine the critical value:
To determine the critical value, we need to consult the t-distribution table with n1 + n2 - 2 degrees of freedom at the given significance level (α = 0.01). Looking up the table, we find the critical value to be approximately ±2.797.
Step 6: Compare the test statistic with the critical value:
If the absolute value of the test statistic is greater than the critical value, we reject the null hypothesis; otherwise, we fail to reject the null hypothesis.
In this case, since the test statistic is less than the critical value (|t| < 2.797), we fail to reject the null hypothesis. Therefore, we can regard both machines as equally efficient at the 1% level of significance.