No, there is insufficient evidence to conclude that the mean fill volume differs from 10 ounces.
To perform a hypothesis test to determine if the mean fill volume is greater than 10 ounces, we will follow these steps:
State the null and alternative hypotheses:
Null hypothesis (H₀): The mean fill volume is equal to 10 ounces. (μ = 10)
Alternative hypothesis (H₁): The mean fill volume is greater than 10 ounces. (μ > 10)
Determine the significance level (α): α = 0.10
Calculate the test statistic:
We will use a one-sample t-test since the population standard deviation is unknown. The test statistic is calculated as:
t = (x - μ₀) / (s / √n)
where:
x is the sample mean (10.11125 ounces)
μ₀ is the hypothesized mean (10 ounces)
s is the sample standard deviation (0.03871 ounces)
n is the sample size (8 cans)
Plugging in the values, we get:
t = (10.11125 - 10) / (0.03871 / √8) ≈ 2.68
Find the p-value:
The p-value is the probability of obtaining a test statistic as extreme or more extreme than the one calculated, assuming the null hypothesis is true. We can find the p-value using a t-distribution table or statistical software. For a one-tailed test with α = 0.10 and degrees of freedom (df) = n - 1 = 7, the p-value is approximately 0.015.
Make a decision:
Since the p-value (0.015) is less than the significance level (α = 0.10), we reject the null hypothesis. This means that there is sufficient evidence to conclude that the mean fill volume is greater than 10 ounces.
Therefore, we can conclude that the mean fill volume is greater than 10 ounces at the 10% level of significance.