Annual per capita consumption of milk is 21.6 gallons (Statistical Abstract of the United States: 2006). Being from the Midwest, you believe milk consumption is higher there and wish to support your opinion. A sample of 16 individuals from the mid western town of Webster City showed a sample mean annual consumption of 24.1 gallons with a standard deviation of s = 4.8.
1. What is a point estimate of the difference between mean annual consumption in Webster City and the national mean?
2. At = .05, test for a significant difference by completing the following.
Calculate the value of the test statistic (to 2 decimals).
3. The p-value is?
1. To calculate the point estimate of the difference between mean annual consumption in Webster City and the national mean, subtract the national mean from the sample mean in Webster City:
Point Estimate = Sample Mean - National Mean
Point Estimate = 24.1 gallons - 21.6 gallons
Point Estimate = 2.5 gallons
The point estimate of the difference is 2.5 gallons.
2. To test for a significant difference, we can use a t-test. The formula for the t-test statistic is:
t = (Sample Mean - Population Mean) / (Sample Standard Deviation / √(Sample Size))
Here, the sample mean is 24.1, the population mean is 21.6, the sample standard deviation is 4.8, and the sample size is 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 value of the test statistic is approximately 2.08.
3. To find the p-value, you need to consult a t-distribution table or use statistical software. The p-value is the probability of getting a test statistic as extreme as the one observed, assuming the null hypothesis is true. In this case, the null hypothesis would be that there is no significant difference between mean annual consumption in Webster City and the national mean.
For a two-tailed test at α = 0.05, the critical value (rejection region) is ±2.131 from the t-distribution table (since the degrees of freedom is n-1 = 16-1 = 15).
As the calculated test statistic (t = 2.08) does not fall beyond the rejection region, we do not reject the null hypothesis.
The p-value can be calculated from the t-distribution table or using statistical software. In this case, the p-value would be greater than 0.10 (since the test statistic does not exceed the critical value of 2.131).
Therefore, the p-value is greater than 0.10.