A packaging device is set to fill detergent power packets 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 we calculate that the mean weight produced by the machine has increased? Use 5% level of significance.
Well, it seems like the detergent power packets are feeling a bit heavyweight lately. Let's see if the increase is statistically significant.
To test whether the mean weight produced by the machine has increased, we can use a one-sample t-test. The null hypothesis (H0) would be that the mean weight is still 5 Kg, while the alternative hypothesis (H1) would be that the mean weight has increased.
Now, let me put on my mathematician wig here and crunch some numbers. The formula for the t-statistic is:
t = (sample mean - population mean) / (sample standard deviation / sqrt(sample size))
Plugging in the numbers, we get:
t = (5.03 - 5) / (0.21 / sqrt(100))
Simplifying that, we get:
t = 0.03 / (0.21 / 10)
Now, calculating that, we find:
t ≈ 1.43
Next, we need to compare this t-value to the critical t-value at a 5% level of significance. Looking up in our trusty statistical tables, we find that for a one-tailed test (since we want to know if it has increased), with degrees of freedom (n-1) equal to 99, the critical t-value is approximately 1.66.
Since the calculated t-value of 1.43 is less than the critical t-value of 1.66, we fail to reject the null hypothesis. In simpler terms, we cannot confidently say that the mean weight produced by the machine has increased.
So, it seems like the packets are not getting any heavier. Perhaps they just hit the weights at the gym.
To determine if the mean weight produced by the machine has increased, we can conduct a hypothesis test using the given sample data.
Let's define the null hypothesis (H0) and the alternative hypothesis (Ha):
H0: The mean weight produced by the machine has not increased (μ = 5 Kg)
Ha: The mean weight produced by the machine has increased (μ > 5 Kg)
To test this hypothesis, we can use a one-sample t-test, as we have the sample mean and standard deviation.
Here's how we can perform the hypothesis test:
1. Set the significance level (α) to 0.05 (5% level of significance).
2. Calculate the test statistic, t-score, using the formula:
t = (sample mean - population mean) / (sample standard deviation / √n)
Where:
Sample mean = 5.03 Kg
Population mean = 5 Kg
Sample standard deviation = 0.21 Kg
n = sample size = 100
t = (5.03 - 5) / (0.21 / √100)
t = 0.03 / (0.21 / 10)
t = 1.43
3. Determine the degrees of freedom, which is equal to the sample size minus 1. In this case, the degrees of freedom (df) = 100 - 1 = 99.
4. Find the critical t-value for a one-tailed test with a significance level of 0.05 and degrees of freedom of 99. This critical t-value can be found using statistical tables or calculators, and in this case, it is approximately 1.66.
5. Compare the calculated t-score with the critical t-value.
Since the calculated t-score (1.43) is less than the critical t-value (1.66), we fail to reject the null hypothesis. This means that we do not have sufficient evidence to conclude that the mean weight produced by the machine has increased.
In simpler terms, based on the given sample data, we cannot say with 95% confidence that the mean weight produced by the machine has increased. The observed increase of 0.03 Kg could be due to random variation or measurement error.
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X-x/q
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.