a packaging device is set to fill detergent powder packets with a mean weight of 5kg.The standard deviation is known to be .01kg.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 weighted.this sample has a mean of 5.03kg and a standard deviation of .21kg.can we calculate that the mean weight produced by the machine has increased.use 5% level of significance?
4n /8.8
To determine if the mean weight produced by the machine has increased, we can conduct a hypothesis test.
Null Hypothesis (H₀): The mean weight produced by the machine has not increased.
Alternative Hypothesis (H₁): The mean weight produced by the machine has increased.
We can conduct a one-sample t-test to compare the sample mean to the hypothesized population mean.
Given:
Hypothesized population mean (μ₀) = 5 kg
Sample mean (x̄) = 5.03 kg
Sample size (n) = 100
Sample standard deviation (s) = 0.21 kg
Level of significance (α) = 0.05
Next, we calculate the t-statistic using the formula:
t = (x̄ - μ₀) / (s / √n)
t = (5.03 - 5) / (0.21 / √100)
t = 0.03 / (0.21 / 10)
t = 0.03 / 0.021
t ≈ 1.4286
Now, we need to find the critical value (t-critical) using the significance level (α) and degrees of freedom (n-1).
Since the sample size is 100, degrees of freedom = 100 - 1 = 99.
Using a t-table or software, we find the t-critical value for a one-tailed test with α = 0.05 and df = 99 is approximately 1.660.
Since the calculated t-statistic (1.4286) < t-critical (1.660), we fail to reject the null hypothesis.
Therefore, based on the given data, we do not have sufficient evidence to conclude that the mean weight produced by the machine has increased at a 5% level of significance.
To determine if the mean weight produced by the machine has increased, we can conduct a hypothesis test using the sample data provided. Here are the steps to calculate the test statistic and make a conclusion:
Step 1: State the null and alternative hypotheses:
Null Hypothesis (H0): The mean weight produced by the machine has not increased.
Alternative Hypothesis (H1): The mean weight produced by the machine has increased.
Step 2: Determine the level of significance:
The level of significance is given as 5% or 0.05.
Step 3: Calculate the test statistic:
The test statistic in this case is the z-score, which measures how many standard deviations the sample mean is away from the hypothesized population mean. We can calculate the z-score using the formula:
z = (x̄ - μ) / (σ / sqrt(n))
Where:
x̄ is the sample mean (5.03kg),
μ is the hypothesized population mean (5kg),
σ is the standard deviation of the population (.01kg),
n is the sample size (100).
Plugging in the values, we get:
z = (5.03 - 5) / (0.01 / sqrt(100))
z = 0.03 / (0.01 / 10)
z = 3
Step 4: Determine the critical value:
Since we are using a 5% level of significance, we need to compare the z-score to the critical value of a standard normal distribution at alpha = 0.05. The critical value for a one-tailed test is approximately 1.645.
Step 5: Make a decision:
If the calculated test statistic (z-score) is greater than the critical value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
In this case, the calculated test statistic (z = 3) is greater than the critical value (1.645), which means it falls in the rejection region. Therefore, we can conclude that the mean weight produced by the machine has increased at a 5% level of significance.
Please note that this conclusion assumes basic conditions are met, such as a random sample, normality of the data, and independence of observations.
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n
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.