The international Coffee Association has reported the mean daily coffee consumption for U.S. residents as 1.65 cups. Assume that a sample of 38 people from a North Carolina city consumed a mean of 1.84 cups of coffee per day, with a standard deviation of 0.85 cups. In a two-tail test at the 0.05 level, could the residents of this city be said to be significantly different from their counterparts across the nation?
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. Is Z > 1.96?
To determine if the residents of this city are significantly different from their counterparts across the nation, we can conduct a hypothesis test.
Step 1: State the null and alternative hypotheses:
Null Hypothesis (H0): The mean daily coffee consumption for the residents of this city is not significantly different from the national mean.
Alternative Hypothesis (Ha): The mean daily coffee consumption for the residents of this city is significantly different from the national mean.
Step 2: Set the significance level (alpha):
The significance level (alpha) is given as 0.05.
Step 3: Determine the test statistic:
Since the population standard deviation is unknown, we will use a t-test. The test statistic is calculated using the formula:
t = (sample mean - population mean) / (sample standard deviation / square root of sample size)
In this case:
Sample mean (x̄) = 1.84 cups
Population mean (μ) = 1.65 cups
Sample standard deviation (s) = 0.85 cups
Sample size (n) = 38
Plugging in the values:
t = (1.84 - 1.65) / (0.85 / √38) = 2.22
Step 4: Determine the critical value(s):
Since it is a two-tail test, the critical values are obtained from the t-distribution with n-1 degrees of freedom and a significance level of 0.05. In this case, the degrees of freedom are 38-1 = 37.
Using a t-table or a statistical software, the critical values are approximately t = ±2.024.
Step 5: Make a decision:
If the test statistic falls outside the critical value range, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
In this case, the test statistic is 2.22, which is greater than the positive critical value of 2.024. Therefore, we reject the null hypothesis.
Step 6: State the conclusion:
Based on the evidence, we can conclude that the residents of this city have a significantly different mean daily coffee consumption compared to their counterparts across the nation, at the 0.05 significance level.
To determine whether the residents of the North Carolina city are significantly different from the counterparts across the nation in terms of coffee consumption, we can perform a two-tail hypothesis test using the given data.
Let's define the null and alternative hypotheses:
Null Hypothesis (H0): The mean coffee consumption in the North Carolina city is equal to the mean coffee consumption reported by the international Coffee Association (μ = 1.65 cups).
Alternative Hypothesis (Ha): The mean coffee consumption in the North Carolina city is significantly different from the mean coffee consumption reported by the international Coffee Association (μ ≠ 1.65 cups).
Next, we can calculate the test statistic, which is the z-score, using the formula:
z = (x - μ) / (σ / √n)
where:
x = sample mean (1.84 cups)
μ = population mean (1.65 cups)
σ = population standard deviation (0.85 cups)
n = sample size (38)
Plugging in the values, we get:
z = (1.84 - 1.65) / (0.85 / √38) ≈ 2.42
To determine if the North Carolina city residents are significantly different from the counterparts across the nation, we need to compare the calculated z-score with the critical z-value at a 0.05 significance level (two-tailed test).
The critical z-value can be found using a z-table or a statistical calculator. For a two-tailed test with a significance level of 0.05, the critical z-value is approximately ±1.96.
Since the calculated z-score (2.42) falls outside the critical region (±1.96), we can reject the null hypothesis in favor of the alternative hypothesis. This means that the residents of the North Carolina city can be said to be significantly different from their counterparts across the nation in terms of coffee consumption.