The average number of articles produce by two machines 200 and 250 with standard deviation 20 and 25 respectively on the basis of records of 25 days production. Can you regard both the machine equally efficient at 1% level of significance.
Well, let's see if these machines pass the efficiency test or if they're just clowning around!
To determine if both machines are equally efficient, we can perform a hypothesis test. Our null hypothesis (H0) would be that the average number of articles produced by both machines is equal, while the alternative hypothesis (H1) would be that they are not equal.
At the 1% level of significance, we'll need to calculate the critical value for a two-tailed z-test. Since we have two machines, we'll be comparing two sample means.
The formula for the test statistic (z) is:
z = (x1 - x2) / √((σ1^2 / n1) + (σ2^2 / n2))
Where:
x1 and x2 are the sample means (200 and 250)
σ1 and σ2 are the standard deviations (20 and 25)
n1 and n2 are the sample sizes (25 each)
Calculating the test statistic, we get:
z = (200 - 250) / √((20^2 / 25) + (25^2 / 25))
z = -50 / √(400 + 625)
z = -50 / √1025
z ≈ -50 / 32.02
z ≈ -1.56
Now, let's check if the test statistic falls within the critical region. At the 1% level of significance, the critical value is approximately ±2.57 (two-tailed test).
Since |-1.56| < 2.57, we fail to reject the null hypothesis. In other words, we have insufficient evidence to conclude that the machines are not equally efficient at the 1% level of significance.
So, according to this analysis, we can consider both machines as equally efficient. But remember, this is just statistical clowning - actual machine performance may vary!
To determine if both machines are equally efficient at a 1% level of significance, we can perform a two-sample t-test.
Step 1: State the null and alternative hypotheses:
Null Hypothesis (H0): The average number of articles produced by machine 200 is equal to the average number of articles produced by machine 250.
Alternative Hypothesis (Ha): The average number of articles produced by machine 200 is not equal to the average number of articles produced by machine 250.
Step 2: Calculate the test statistic:
The test statistic for a two-sample t-test is given by the formula:
t = (x1 - x2) / sqrt((s1^2 / n1) + (s2^2 / n2))
where:
x1 and x2 are the sample means,
s1 and s2 are the sample standard deviations,
n1 and n2 are the sample sizes.
Let's calculate the test statistic:
x1 = 200, x2 = 250
s1 = 20, s2 = 25
n1 = n2 = 25 (as stated)
t = (200 - 250) / sqrt((20^2 / 25) + (25^2 / 25))
Step 3: Determine the critical value:
At a 1% level of significance and a two-tailed test, the critical value for the t-test with 48 degrees of freedom is approximately ±2.714.
Step 4: 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.
Step 5: Make a conclusion:
Compare the absolute value of the test statistic with the critical value:
If |t| > 2.714, we reject the null hypothesis that both machines are equally efficient.
If |t| ≤ 2.714, we fail to reject the null hypothesis.
Note: We have not been provided with the actual values for the number of articles produced by each machine, so we cannot calculate the test statistic and make a conclusion.
To determine if both machines are equally efficient at the 1% level of significance, we can use a hypothesis test.
Let's denote the average number of articles produced by machine 200 as μ1 and the average number of articles produced by machine 250 as μ2.
The null hypothesis (H0) is that both machines are equally efficient, meaning that μ1 = μ2. The alternative hypothesis (Ha) is that the machines are not equally efficient, meaning that μ1 ≠ μ2.
To conduct the hypothesis test, we will use a two-sample t-test, assuming that the population variances are unknown but equal. Here are the steps to calculate the test statistic:
1. Calculate the sample mean for each machine:
- Machine 200: x̄1 = 200
- Machine 250: x̄2 = 250
2. Calculate the sample standard deviation for each machine:
- Machine 200: s1 = 20
- Machine 250: s2 = 25
3. Calculate the pooled standard deviation (sp) using the formula:
sp = sqrt(((n1-1)s1^2 + (n2-1)s2^2) / (n1 + n2 - 2))
In this case, n1 = n2 = 25.
4. Calculate the test statistic t using the formula:
t = (x̄1 - x̄2) / (sp * sqrt(1/n1 + 1/n2))
5. Determine the degrees of freedom for the test statistic using the formula:
df = n1 + n2 - 2
6. Look up the critical t-value for the desired level of significance and the degrees of freedom. For a 1% level of significance and 48 degrees of freedom, the critical t-value is approximately 2.677.
7. Compare the absolute value of the test statistic to the critical t-value. If the absolute value of the test statistic is greater than the critical t-value, we reject the null hypothesis; otherwise, we fail to reject the null hypothesis.
If the null hypothesis is rejected, it would imply that the machines are not equally efficient at the 1% level of significance.
the average number of articles produced by two machines per day are 200 and 250 with standard deviation of 20 and 25 respectively on the basis of 25 days production. Can you regard both the machines equally efficient at 1% level of significance?
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n (n = 25)
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 the Z score.