1. To find the point estimate of the difference between the mean annual consumption in Webster City and the national mean, you need to subtract the national mean from the sample mean.
Point Estimate = Sample Mean - National Mean
Point Estimate = 24.1 - 21.6
Point Estimate = 2.5 gallons
The point estimate of the difference between mean annual consumption in Webster City and the national mean is 2.5 gallons.
2. To test for a significant difference, you can perform a hypothesis test.
Null Hypothesis (H0): There is no significant difference between the mean annual consumption in Webster City and the national mean.
Alternative Hypothesis (Ha): There is a significant difference between the mean annual consumption in Webster City and the national mean.
Since the sample size is small (n=16) and the population standard deviation (σ) is unknown, you can use the t-distribution to calculate the test statistic.
The formula for the t-test statistic is:
t = (x̄ - μ) / (s / √n)
where x̄ is the sample mean, μ is the population mean (in this case, the national mean), s is the sample standard deviation, and n is the sample size.
Plugging in the values:
x̄ = 24.1
μ = 21.6
s = 4.8
n = 16
t = (24.1 - 21.6) / (4.8 / √16)
t = 2.5 / (4.8 / 4)
t = 2.5 / 1.2
t ≈ 2.08 (rounded to 2 decimal places)
The test statistic value is approximately 2.08.
3. To find the p-value, you need to compare the test statistic value to the t-distribution table or use statistical software.
Since the significance level (α) is given as 0.05, you are testing at a 95% confidence level. This means that you are looking for a two-tailed test and need to find the critical t-value that corresponds to an alpha level of 0.025 (0.05 / 2) in each tail.
Using the t-distribution table or statistical software, you can find the critical t-value. Let's assume it is t_critical = 2.145 for a degree of freedom (df) of 15 (n-1).
If the absolute value of the test statistic value (|t|) is greater than the critical t-value (|t_critical|), then the p-value is less than the significance level (α), and you reject the null hypothesis. Otherwise, you fail to reject the null hypothesis.
In this case, |t| = 2.08, which is less than |t_critical| = 2.145. Therefore, the p-value is greater than 0.05, and we fail to reject the null hypothesis.
Unfortunately, without the exact distribution of the sample, we cannot determine the exact p-value. However, since the p-value is greater than 0.05, it suggests that there is not enough evidence to support a significant difference in milk consumption between Webster City and the national mean at the 95% confidence level.