a packing device is set to fill detergent powder packet with a mean weight of 5 kg . The standard deviation is known to be 0.01 kg. these are known to drift upwards over a period of time due to machine fault, which is not tolerable. A random sample of 100 packets is taken and weighed. This sample has a mean weight of 5.03 kg and a standard deviation of 0.21 kg. can you conclude that the mean weight produced by the machine has increased? use a 5% level of significant?
To determine if the mean weight produced by the machine has increased, we can perform a hypothesis test using the sample data.
1. State the Null Hypothesis (H0): The mean weight produced by the machine has not increased.
H0: μ = 5 kg (mean weight is 5 kg)
2. State the Alternative Hypothesis (H1): The mean weight produced by the machine has increased.
H1: μ > 5 kg (mean weight is greater than 5 kg)
3. Select the Significance Level (α): The question states to use a 5% level of significance, which means α = 0.05.
4. Calculate the Test Statistic: We will use the Z-test, which is appropriate when the sample size is large (n > 30) and the population standard deviation (known in this case) is given.
The test statistic is given by: Z = (x̄ - μ) / (σ / √n)
where x̄ is the sample mean, μ is the assumed population mean, σ is the population standard deviation, and n is the sample size.
Given:
x̄ = 5.03 kg (sample mean)
μ = 5 kg (mean weight assumed by the machine)
σ = 0.01 kg (population standard deviation)
n = 100 (sample size)
The test statistic is: Z = (5.03 - 5) / (0.01 / √100) = 3
5. Determine the Critical Value: Since the alternative hypothesis is one-tailed (mean weight has increased), we need to find the critical value corresponding to the significance level.
At a 5% level of significance (α = 0.05) and given that this is a one-tailed test:
The critical value can be obtained from the Z-table or using a calculator and is approximately 1.645.
6. Compare the Test Statistic and Critical Value: If the test statistic is greater than the critical value, we reject the null hypothesis and conclude that the mean weight produced by the machine has increased.
In this case, Z (test statistic) = 3 > 1.645 (critical value).
7. Make a Conclusion: Since the test statistic is greater than the critical value, we can reject the null hypothesis and conclude that there is sufficient evidence to suggest that the mean weight produced by the machine has increased.
Therefore, based on the given data and using a 5% level of significance, we can conclude that the mean weight produced by the machine has increased.
To determine if the mean weight produced by the machine has increased, we can perform a hypothesis test.
Step 1: Define the null and alternative hypotheses:
Null Hypothesis (H0): The mean weight produced by the machine has not increased.
Alternative Hypothesis (Ha): The mean weight produced by the machine has increased.
Step 2: Determine the significance level:
The significance level, denoted as α, is given as 5%, which can also be written as 0.05.
Step 3: Calculate the test statistic:
We will use a one-sample t-test because the sample size (n = 100) is over 30.
t = (sample mean - population mean) / (sample standard deviation / sqrt(n))
t = (5.03 - 5) / (0.21 / sqrt(100))
t = 0.03 / (0.21 / 10)
t = 0.03 / 0.021
t ≈ 1.43
Step 4: Determine the critical value:
Since the sample size is large, we can use a Z-table to find the critical value at a 5% significance level. The critical value for a one-tailed test is approximately 1.645.
Step 5: Compare the test statistic with the critical value:
Since the test statistic (1.43) is less than the critical value (1.645), we fail to reject the null hypothesis.
Step 6: Make the conclusion:
Based on the test results, we do not have sufficient evidence to conclude that the mean weight produced by the machine has increased at a 5% level of significance.