The U.S. Department of Agriculture claims that the mean consumption of coffee by the person in the United States is 24.2 gallons per year. A random sample of 120 people in the United States shows that the mean coffee consumption is 23.5 gallons per year with a standard deviation of 3.2 gallons. At a=0.05, can you reject the claim?

Try a one-sample z-test.

Hypotheses:
Ho: µ = 24.2 -->null hypothesis
Ha: µ does not equal 24.2 -->alternate hypothesis

Check the cutoff or critical value to reject the null using a z-table at 0.05 level for a two-tailed test. If the test statistic calculated exceeds either the positive or negative critical value from the table, reject the null and conclude a difference. If the test statistic does not exceed the critical value from the table, do not reject the null.

Use 23.5 as the sample mean.
Use 3.2 for standard deviation.
Use 120 for sample size.

I hope this will help get you started.

To determine if we can reject the claim made by the U.S. Department of Agriculture, we need to conduct a hypothesis test.

Step 1: State the null and alternative hypotheses.
Null Hypothesis (H₀): The mean coffee consumption in the United States is equal to 24.2 gallons per year.
Alternative Hypothesis (H₁): The mean coffee consumption in the United States is not equal to 24.2 gallons per year.

Step 2: Formulate the test statistic.
We will use the t-test statistic since the population standard deviation is unknown and the sample size is small (n < 30). The formula for the t-test statistic is:
t = (x̄ - μ) / (s / √n)
where x̄ = sample mean, μ = population mean, s = sample standard deviation, and n = sample size.

Step 3: Set the significance level.
The significance level (α) is given as 0.05, which indicates a 5% level of significance.

Step 4: Calculate the test statistic.
t = (23.5 - 24.2) / (3.2 / √120)
t = -0.7 / (3.2 / √120)

Step 5: Determine the critical value.
Since we have a two-tailed test (H₁: μ ≠ 24.2), we need to find the critical t-value for α/2 = 0.025 and degrees of freedom (df) = n - 1.
df = 120 - 1 = 119
Using a t-table or calculator, we find that the critical t-value (tₐ) is approximately ±1.980.

Step 6: 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.

|t| > tₐ ?
|t| > 1.980 ?
|(-0.7 / (3.2 / √120))| > 1.980 ?
(0.7 / (3.2 / √120)) > 1.980 ?

Step 7: Calculate and interpret the p-value.
To calculate the p-value, we would need the t-distribution. Based on the calculated test statistic, we can determine the corresponding p-value.

P-value > α ?
P-value > 0.05 ?

If the p-value is greater than the significance level (α), which in this case is 0.05, then we fail to reject the null hypothesis.

Please note that the final step (Step 7) requires the t-distribution to determine the p-value accurately. Since this platform does not have the capability to look up values on the t-distribution, I am unable to provide the exact p-value in this case. You can utilize a statistical calculator or software that provides the p-value based on the t-distribution to find the final decision.

To determine whether you can reject the claim made by the U.S. Department of Agriculture, you need to perform a hypothesis test. This involves setting up null and alternative hypotheses, calculating a test statistic, and comparing it to a critical value.

Let's define the null hypothesis (H0) as the claim made by the U.S. Department of Agriculture, which states that the mean consumption of coffee by the person in the United States is 24.2 gallons per year. The alternative hypothesis (Ha) would be that the mean consumption is different from 24.2 gallons per year.

H0: μ = 24.2 gallons
Ha: μ ≠ 24.2 gallons

Next, we need to calculate the test statistic. In this case, we'll use the t-test since the population standard deviation is unknown, and we have a sample size of 120.

The formula for the t-test statistic is:
t = (sample mean - hypothesized mean) / (sample standard deviation / √sample size)

Plugging in the values:
t = (23.5 - 24.2) / (3.2 / √120)

Calculating this gives us:
t = -0.7 / (3.2 / √120)
t ≈ -0.7 / (3.2 / 10.9545)
t ≈ -0.7 / 0.2915
t ≈ -2.4

Now, we need to compare this t-value to the critical value at a significance level of α = 0.05. Since it is a two-tailed test (Ha: μ ≠ 24.2 gallons), we will split the significance level in half (α/2 = 0.025) for each tail.

Looking up the critical value in a t-distribution table with 120 - 1 = 119 degrees of freedom and a significance level of 0.025, we find that the critical value is approximately ±1.980.

Since |-2.4| > 1.980, we can reject the null hypothesis.

Therefore, based on the given data and using a significance level of 0.05, we have enough evidence to suggest that the mean coffee consumption by the person in the United States is different from 24.2 gallons per year.