A brand of powdered milk is advertised as having a net weight of 150 grams. A curious consumer obtained the net weight of 10 randomly selected cans. The values obtained are 256, 248, 242, 245, 246, 248, 250, 255, 243, and 149 grams. Is there a reason to believe that the average net weight of the powdered milk cans is less than 250 grams at 10% level of significance? Assume the net weight is normally distributed with unknown population variance.
To determine if there is reason to believe that the average net weight of the powdered milk cans is less than 250 grams at the 10% level of significance, we will conduct a one-sample t-test.
Step 1: State the null and alternative hypotheses
The null hypothesis (H0) is that the average net weight of powdered milk cans is equal to 250 grams. The alternative hypothesis (H1) is that the average net weight is less than 250 grams.
H0: μ = 250
H1: μ < 250
Step 2: Calculate the test statistic
First, we need to calculate the sample mean and sample standard deviation.
Sample mean (x̄) = (256+248+242+245+246+248+250+255+243+149) / 10 = 2433 / 10 = 243.3 grams
Now, we find the sample variance:
Variance: Σ(xi−x̄)²/(N−1)
= (3.3^2 + 2^2 + 1.3^2 + 4.7^2 + 0^2) / 9
= (10.89 + 4 + 1.69 + 22.09 + 0) / 9
= 38.67 / 9
= 4.3
Sample standard deviation: √variance = √4.3 = 2.07
Now we can calculate the test statistic:
t = (x̄ - μ) / (s / √n)
t = (243.3 - 250) / (2.07 / √10)
t = -6.7 / (2.07 / 3.16)
t = -6.7 / 0.66
t ≈ -10.15
Step 3: Determine the critical value and p-value
We are given a 10% level of significance. Since this is a one-tailed test (we're only interested if the average weight is less than 250 grams), the critical value for a t-distribution with 9 degrees of freedom (sample size - 1) is -1.383 (rounded to three decimal places) [Reference: t-distribution table or calculator].
Using a t-table or calculator, we find the p-value associated with the test statistic -10.15. Given that our test statistic is much lower than -1.383, we can conclude that the p-value will be much smaller than 0.1 (10% level of significance).
Step 4: Compare the test statistic to the critical value and evaluate the p-value
Since our test statistic (-10.15) is less than the critical value (-1.383), we reject the null hypothesis. The p-value is also less than the level of significance (0.1), which further supports the rejection of the null hypothesis.
Conclusion:
There is reason to believe that the average net weight of the powdered milk cans is less than 250 grams at a 10% level of significance.
To determine if there is a reason to believe that the average net weight of the powdered milk cans is less than 250 grams at a 10% level of significance, we will conduct a t-test for a single sample mean. Here are the steps to follow:
Step 1: Null and alternative hypothesis
- Null hypothesis (H0): The average net weight of the powdered milk cans is 250 grams.
- Alternative hypothesis (Ha): The average net weight of the powdered milk cans is less than 250 grams.
Step 2: Calculate the sample mean and sample standard deviation
The sample mean can be calculated by summing all the net weights and dividing by the number of samples:
Sample mean (x̄) = (256 + 248 + 242 + 245 + 246 + 248 + 250 + 255 + 243 + 149) / 10
The sample standard deviation (s) can be calculated using the formula:
s = sqrt((Σ(xi - x̄)^2) / (n-1))
Step 3: Compute the test statistic
The t-test statistic can be calculated using the formula:
t = (x̄ - μ) / (s / sqrt(n))
Where:
- x̄ is the sample mean
- μ is the assumed population mean under the null hypothesis (250g)
- s is the sample standard deviation
- n is the number of samples (10 in this case)
Step 4: Determine the critical value
Using a t-distribution table or a statistical software, find the critical t-value corresponding to a one-tailed test at a 10% level of significance, with 9 degrees of freedom.
Step 5: Compare the t-value with the critical value
If the calculated t-value is less than the critical value, we reject the null hypothesis.
Follow these steps to calculate the t-value and determine if there is a reason to believe that the average net weight of the powdered milk cans is less than 250 grams at a 10% level of significance.
To determine whether there is a reason to believe that the average net weight of the powdered milk cans is less than 250 grams at a 10% level of significance, you can conduct a one-sample t-test. This statistical test will help you determine if the sample mean significantly differs from a specified value (in this case, 250 grams).
Here is how you can perform the one-sample t-test:
Step 1: Set up hypotheses
- Null hypothesis (H0): The average net weight of the powdered milk cans is equal to or greater than 250 grams.
- Alternative hypothesis (Ha): The average net weight of the powdered milk cans is less than 250 grams.
Step 2: Calculate the sample mean and standard deviation
- Calculate the mean (x̄) of the 10 net weights obtained: (256 + 248 + 242 + 245 + 246 + 248 + 250 + 255 + 243 + 149) / 10 = 244.2 grams.
- Calculate the sample standard deviation (s) of the net weights using the formula: √[Σ(xi - x̄)² / (n - 1)], where xi represents each net weight and n is the sample size.
Step 3: Calculate the test statistic
- Calculate the test statistic using the formula: t = (x̄ - μ0) / (s / √n), where μ0 is the specified value in the null hypothesis (250 grams), x̄ is the sample mean, s is the sample standard deviation, and n is the sample size.
Step 4: Determine the critical value
- With a 10% level of significance and 9 degrees of freedom (n - 1), calculate the critical value from the t-distribution table or using statistical software.
Step 5: Compare the test statistic with the critical value
- If the test statistic is less than the critical value, reject the null hypothesis. Otherwise, fail to reject the null hypothesis.
Performing these steps will help you determine whether there is evidence to support the claim that the average net weight of the powdered milk cans is less than 250 grams at a 10% level of significance.