a potato chip manufacturer has been accused of shortchanging their customers with 10 ouce of bags potato chips. a sample of 50 of that company's 10 ounce bags was take, and it was found that the average weight was 9.8 ounces with a standard deviation of 0.3 ounces.can we conclude that the company's bags of potato chips are less than 10 ounces?use 5%significance level
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.
To determine if the company's bags of potato chips are less than 10 ounces, we can perform a hypothesis test. The null hypothesis, denoted as H0, is that the average weight of the bags is equal to 10 ounces. The alternative hypothesis, denoted as Ha, is that the average weight of the bags is less than 10 ounces.
1. Set up the hypotheses:
- Null hypothesis (H0): The average weight of the bags = 10 ounces.
- Alternative hypothesis (Ha): The average weight of the bags < 10 ounces.
2. Determine the significance level, denoted as α (alpha). In this case, it is given as 5%, which means α = 0.05.
3. Calculate the test statistic. Since we have sample data and the population standard deviation is unknown, we will use a t-test. The formula for the t-test statistic is:
t = (sample mean - population mean) / (sample standard deviation / sqrt(sample size))
Given:
sample mean (x̄) = 9.8 ounces
population mean (μ) = 10 ounces
sample standard deviation (s) = 0.3 ounces
sample size (n) = 50
Plugging in the values:
t = (9.8 - 10) / (0.3 / sqrt(50))
4. Calculate the degrees of freedom. For a t-test with a sample size of 50, the degrees of freedom are (n - 1) = (50 - 1) = 49.
5. Determine the critical t-value. Since we are performing a one-tailed test and the test statistic is t < 0, we need to find the critical t-value corresponding to a left-tailed test at a 5% significance level with 49 degrees of freedom. Using a t-table or statistical software, the critical t-value is -1.676.
6. Compare the test statistic with the critical t-value. If the test statistic is less than the critical t-value, we reject the null hypothesis; otherwise, we fail to reject the null hypothesis.
If t < -1.676, we reject H0. Otherwise, we fail to reject H0.
7. Calculate the p-value associated with the test statistic. The p-value represents the probability of observing a test statistic as extreme as the one calculated, assuming the null hypothesis is true. If the p-value is smaller than the significance level (α), we reject the null hypothesis.
Using statistical software or a t-table, we can find the p-value associated with the test statistic.
8. Compare the p-value with the significance level. If the p-value is smaller than the significance level (α < 0.05), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
If p-value < α (0.05), we reject H0. Otherwise, we fail to reject H0.
By following these steps, you should be able to determine whether the company's bags of potato chips are less than 10 ounces based on the sample data.