the typical
American coffee drinker consumes an average of 3.1 cups per day. A sample of 12 senior citizens reveled
they consumed the following amounts, reported in cups, of coffee yesterday.
3.1 3.3 3.5 2.6 2.6 4.3 4.4 3.8 3.1 4.1 3.1 3.2
At the .05 significance level, does this sample data suggest there is a difference between the national
average and the sample information from senior citizens?
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√(n-1)
Since only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z score.
To determine if there is a significant difference between the national average and the sample information from senior citizens, we can perform a t-test.
Here are the step-by-step procedures to conduct the t-test:
Step 1: State the null and alternative hypotheses:
- Null hypothesis (H0): There is no difference between the national average and the sample information from senior citizens.
- Alternative hypothesis (Ha): There is a difference between the national average and the sample information from senior citizens.
Step 2: Select the appropriate test and significance level:
Since we are comparing means, we will use a one-sample t-test. The significance level (alpha) is given as 0.05.
Step 3: Calculate the test statistic:
To calculate the t-test statistic, we need to find the sample mean and standard deviation of the sample data.
Sample mean (x̄) = (3.1 + 3.3 + 3.5 + 2.6 + 2.6 + 4.3 + 4.4 + 3.8 + 3.1 + 4.1 + 3.1 + 3.2) / 12 = 37.1 / 12 = 3.092
Sample standard deviation (s) = sqrt(((3.1 - 3.092)^2 + (3.3 - 3.092)^2 + ... + (3.2 - 3.092)^2) / (12 - 1)) = 0.544
t-test statistic = (x̄ - μ) / (s / sqrt(n))
where μ is the national average (3.1), n is the sample size (12), s is the sample standard deviation, and x̄ is the sample mean.
Substituting the values:
t-test statistic = (3.092 - 3.1) / (0.544 / sqrt(12)) = -0.008 / (0.544 / 3.464) ≈ -0.0146
Step 4: Calculate the p-value:
Using the t-distribution table or software, we can find the p-value corresponding to the t-test statistic (-0.0146) and degrees of freedom (12 - 1 = 11).
From the t-distribution table, the p-value is approximately 0.990.
Step 5: Make a decision:
Compare the p-value to the significance level (alpha). If the p-value is less than or equal to alpha (0.05), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
In this case, the p-value (0.990) is greater than the significance level (0.05). Therefore, we fail to reject the null hypothesis.
Step 6: State the conclusion:
Based on the sample data, there is not enough evidence to suggest a difference between the national average and the sample information from senior citizens at the 0.05 significance level.
To determine if there is a difference between the national average of American coffee drinkers and the sample information from senior citizens, we can use a hypothesis test. Specifically, we can perform a one-sample t-test.
Here are the steps to conduct the hypothesis test:
Step 1: State the null hypothesis (H0) and alternative hypothesis (Ha):
- The null hypothesis assumes that there is no difference between the national average and the sample information from senior citizens.
- The alternative hypothesis assumes that there is a difference between the national average and the sample information from senior citizens.
- In this case, we can state the null hypothesis as: The population mean of senior citizens' coffee consumption is equal to the national average (µ = 3.1 cups per day), and the alternative hypothesis as: The population mean of senior citizens' coffee consumption is not equal to the national average (µ ≠ 3.1 cups per day).
Step 2: Select the significance level:
- The significance level (α) is the threshold used to evaluate the strength of evidence against the null hypothesis.
- In this case, the significance level is given as 0.05, which means we want our test to have a 5% chance of rejecting the null hypothesis when it is actually true.
Step 3: Calculate the test statistic:
- We need to calculate the t-value, which measures the difference between the sample mean and the hypothesized population mean in terms of standard error.
- The formula for the t-value is: t = (x̄ - µ) / (s / √n), where x̄ is the sample mean, µ is the population mean (3.1 cups per day), s is the sample standard deviation, and n is the sample size.
- Calculate the sample mean (x̄), sample standard deviation (s), and sample size (n) from the provided data.
Step 4: Determine the critical value:
- The critical value is based on the significance level and the degrees of freedom (df), which is equal to n - 1.
- Find the critical t-value from a t-distribution table or use a t-distribution calculator.
Step 5: Compare the test statistic with the critical value:
- If the absolute value of the test statistic is greater than the critical value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
Step 6: Interpret the results:
- Based on the comparison, conclude whether there is sufficient evidence to suggest a difference between the national average and the sample information from senior citizens.
Note: Since you provided the sample data, I'll calculate the necessary statistics and perform the hypothesis test.
Calculations:
Sample mean (x̄) = (3.1 + 3.3 + 3.5 + 2.6 + 2.6 + 4.3 + 4.4 + 3.8 + 3.1 + 4.1 + 3.1 + 3.2) / 12 = 3.3667 cups per day
Sample standard deviation (s) = √[((3.1 - 3.3667)^2 + (3.3 - 3.3667)^2 + (3.5 - 3.3667)^2 + (2.6 - 3.3667)^2 + (2.6 - 3.3667)^2 + (4.3 - 3.3667)^2 + (4.4 - 3.3667)^2 + (3.8 - 3.3667)^2 + (3.1 - 3.3667)^2 + (4.1 - 3.3667)^2 + (3.1 - 3.3667)^2 + (3.2 - 3.3667)^2) / (12 - 1)] = 0.5893 cups per day
Now, we can proceed to Step 4 and Step 5 to determine the test statistic and compare it with the critical value.