a certain brand of powdered milk is advertised as having a net weight of 250 grams. if the net weights of a random sample of 10 cans are 253, 248, 252, 245, 247, 249, 251, 250, 247, 248 grams. can it be concluded that the average net weight is less than the advertised amount? Use a= 0.05
Find the mean first = sum of scores/number of scores
Subtract each of the scores from the mean and square each difference. Find the sum of these squares. Divide that by the number of scores to get variance.
Standard deviation = square root of variance
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
isn't this a t-test because it has 10 samples only?
my answer is to accept the null hypothesis because i got a computed value of -1.27 and a table value of -1.645
-1.27< -1.645 (Accept the null hypothesis)
Answer pls
Well, it seems like this particular brand of powdered milk is playing a little game of hide and seek with their net weights. Let's try to find out if the average net weight is less than the advertised amount.
To do this, we first need to set up our hypotheses. Our null hypothesis (H0) would be that the average net weight is equal to the advertised amount, and the alternative hypothesis (Ha) would be that the average net weight is less than the advertised amount.
H0: 𝜇 = 250 grams
Ha: 𝜇 < 250 grams
Now, we can calculate the sample mean (x̄) and sample standard deviation (s) from the given data:
x̄ = (253 + 248 + 252 + 245 + 247 + 249 + 251 + 250 + 247 + 248) / 10
≈ 249.0 grams
Next, we calculate the standard error (SE), which is the standard deviation divided by the square root of the sample size:
SE = s / √n
Using the formula, we find that the standard error is approximately equal to:
SE ≈ 1.964 grams
Now, we need to determine the critical value for a one-tailed test with a significance level of 0.05. Since we have a small sample size (n < 30), we'll use a t-distribution. Looking up the critical value in a t-table (or using a calculator), we find that the critical value is approximately -1.833.
Finally, we can calculate the t-statistic using the formula:
t = (x̄ - 𝜇) / SE
t = (249.0 - 250) / 1.964
≈ -0.507
Comparing the calculated t-statistic to the critical value, we can see that -0.507 is greater than -1.833. This means that the calculated t-statistic does not fall within the critical region.
Since the calculated t-statistic is not in the critical region, we fail to reject the null hypothesis. In other words, we cannot conclude that the average net weight is less than the advertised amount.
So, it seems like this brand of powdered milk is just having a little fluctuation with its net weights – a comedy of errors, if you will.
To determine whether it can be concluded that the average net weight is less than the advertised amount, we can perform a hypothesis test using the given sample data.
Let's set up the hypothesis:
Null Hypothesis (H0): The average net weight is equal to the advertised amount.
Alternative Hypothesis (H1): The average net weight is less than the advertised amount.
Next, we need to determine the test statistic and the critical value. Since the sample size is small (n = 10) and the population standard deviation is unknown, we will use a t-test.
Given:
Sample size (n) = 10
Sample mean (x̄) = (253 + 248 + 252 + 245 + 247 + 249 + 251 + 250 + 247 + 248) / 10 = 249.0
Standard deviation (s) = √[Σ(xi - x̄)² / (n-1)] = √[36.0 / 9] = 2.0
Advertised amount (μ) = 250 grams
Significance level (α) = 0.05
Calculate the test statistic:
t = (x̄ - μ) / (s / √n)
= (249.0 - 250) / (2.0 / √10)
= -1.0 / (2.0 / √10)
= -1.0 / (2.0 / 3.162)
= -1.0 / 0.632
≈ -1.583
Next, find the critical value. Since the alternative hypothesis states that the average net weight is less, it is a one-tailed test. With a significance level of 0.05, we need to find the critical value for the left tail of the t-distribution with 10-1 = 9 degrees of freedom.
Using a t-distribution table or a t-distribution calculator, the critical value for a one-tailed test with 9 degrees of freedom and α = 0.05 is approximately -1.833.
Compare the test statistic and the critical value:
Since the test statistic (-1.583) is greater than the critical value (-1.833), we fail to reject the null hypothesis.
Conclusion:
Based on the sample data and the hypothesis test, there is not enough evidence to conclude that the average net weight is less than the advertised amount.