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. The point estimate of the difference between mean annual consumption in Webster City and the national mean can be calculated as the difference between the sample mean and the national mean.
Point Estimate = Sample Mean - National Mean
= 24.1 - 21.6
= 2.5 gallons
Therefore, the point estimate of the difference is 2.5 gallons.
2. To test for a significant difference between the mean annual consumption in Webster City and the national mean, we can use a t-test. The null hypothesis (H0) states that there is no significant difference between the means, while the alternative hypothesis (Ha) states that there is a significant difference.
The test statistic for a t-test is given by:
t = (Sample Mean - National Mean) / (Standard Deviation / √(Sample Size))
t = (24.1 - 21.6) / (4.8 / √16)
t = 2.5 / (4.8 / 4)
t = 2.5 / 1.2
t ≈ 2.08 (to 2 decimal places)
Therefore, the value of the test statistic is approximately 2.08.
3. The p-value can be determined by comparing the test statistic to the critical value at the specified significance level (α). As α is given as 0.05 (5%), we need to find the critical value for a two-tailed test.
By looking up the critical value in the t-distribution table or using statistical software, we find that the critical value for α = 0.05 and degrees of freedom (df) = (n - 1) = (16 - 1) = 15, is approximately ±2.131.
Since the test statistic t = 2.08 is within the range of ±2.131, we fail to reject the null hypothesis.
The p-value is the probability of obtaining a test statistic as extreme as the one observed, assuming the null hypothesis is true. Since the test statistic t = 2.08 is not extreme enough to fall in the rejection region, the p-value is greater than 0.05.
Therefore, the p-value is greater than 0.05.
1. To find the point estimate of the difference between mean annual consumption in Webster City and the national mean, you subtract the national mean from the sample mean.
Point estimate = Sample Mean - National Mean
In this case, the sample mean annual consumption in Webster City is given as 24.1 gallons, and the national mean is given as 21.6 gallons.
Point estimate = 24.1 - 21.6 = 2.5 gallons
Therefore, 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 between the mean annual consumption in Webster City and the national mean, we need to perform a hypothesis test. The null hypothesis (H0) assumes that there is no significant difference between the two means, while the alternative hypothesis (Ha) assumes that there is a significant difference.
H0: μ (population mean) = 21.6
Ha: μ ≠ 21.6
We can calculate the test statistic using the formula:
t = (Sample Mean - Population Mean) / (Standard Deviation / √Sample Size)
Substituting in the given values:
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)
Therefore, the value of the test statistic is approximately 2.08.
3. The p-value is the probability of observing a test statistic that is as extreme as the one calculated (or more extreme) assuming the null hypothesis is true. To find the p-value, we need to compare the test statistic to the critical value(s) at the chosen significance level (α).
Since the p-value is not provided in the question, we need additional information to calculate it. We need the degrees of freedom (df) to look up the critical value(s) in the t-distribution table.
The degrees of freedom for an independent t-test are equal to the sum of the sample sizes minus 2.
In this case, the sample size is 16.
df = 16 - 2 = 14
Once we have the critical value(s) corresponding to our chosen significance level (let's assume α = 0.05, meaning a 5% level of significance), we can compare it to the test statistic to determine the p-value. If the test statistic falls within the critical region, we reject the null hypothesis.
Unfortunately, without the critical value(s), we cannot directly calculate the p-value.