1) A potato chip manufacturer has been accused of shortchanging their customers with their 10-ounce bags of potato chips. A sample of 50 of that company’s 10-ounce bags was taken, 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 a 5% significance level.

To determine whether we can conclude that the company's bags of potato chips are less than 10 ounces, we can perform a hypothesis test. The hypothesis test involves setting up null and alternative hypotheses, calculating a test statistic, and comparing it to a critical value.

Step 1: Set up the hypotheses:
The null hypothesis (H0) is the statement we assume to be true unless there is sufficient evidence to prove otherwise. In this case, the null hypothesis would be that the company's bags of potato chips have an average weight of 10 ounces.

H0: μ = 10 ounces

The alternative hypothesis (Ha) is the statement we're trying to find evidence for. In this case, we want to determine if the company's bags of potato chips have an average weight less than 10 ounces.

Ha: μ < 10 ounces

Step 2: Determine the significance level (alpha):
The significance level, denoted as alpha (α), represents the probability of rejecting the null hypothesis when it's true. In this case, a 5% significance level is given, which means alpha (α) = 0.05.

Step 3: Calculate the test statistic:
The test statistic depends on the distribution of the population and the type of data we're dealing with. Since we have a sample mean and standard deviation, we can use the t-distribution.

The formula for the t-test statistic in this case is:
t = (x̄ - μ) / (s / √n),
where x̄ is the sample mean, μ is the population mean (assumed under the null hypothesis), s is the sample standard deviation, and n is the sample size.

Given:
Sample mean (x̄) = 9.8 ounces
Population mean assumed under H0 (μ) = 10 ounces
Sample standard deviation (s) = 0.3 ounces
Sample size (n) = 50

t = (9.8 - 10) / (0.3 / √50)

Step 4: Determine the critical value:
Since we want to test if the average weight is less than 10 ounces, we need to find the critical value from the t-distribution with a one-tailed test. The critical value marks the boundary beyond which we reject the null hypothesis. We can use a t-table or statistical software to find the critical value.

For a one-tailed test at a 5% significance level and degrees of freedom (df) equal to the sample size minus 1 (df = n - 1 = 50 - 1 = 49), the critical value is approximately -1.68.

Step 5: Compare the test statistic with the critical value:
If the test statistic falls within the critical region, we reject the null hypothesis. In this case, since we're performing a one-tailed test in the lower tail, we reject the null hypothesis if the test statistic is less than the critical value (-1.68).

If the test statistic is less than the critical value, we reject the null hypothesis:
t < -1.68

Step 6: Make a decision:
If the test statistic falls within the critical region (t < -1.68), we reject the null hypothesis. This means there is sufficient evidence to conclude that the company's bags of potato chips are less than 10 ounces.

In summary, we calculate the t-test statistic using the given sample data and compare it to the critical value. If the test statistic is less than the critical value, we reject the null hypothesis and conclude that the company's bags of potato chips are less than 10 ounces.