Based on the available information, there is no sufficient evidence that at a =0.01 we can conclude that their healthcare expenditure differs from the national average of $1468
To determine if there is sufficient evidence to conclude that the healthcare expenditure of the students differs from the national average, we can conduct a hypothesis test.
Let's set up the null and alternative hypotheses:
Null hypothesis (H0): The mean healthcare expenditure of the students is equal to the national average ($1468).
Alternative hypothesis (Ha): The mean healthcare expenditure of the students is different from the national average ($1468).
We can use a t-test since we have the sample mean, sample size, and population standard deviation.
Given:
Sample mean (x) = $1520
Population standard deviation (σ) = $198
Sample size (n) = 60
Significance level (α) = 0.01
To conduct the t-test, we calculate the test statistic using the formula:
t = (x - μ) / (σ / √n)
where x is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size.
In this case, the population mean is $1468.
t = (1520 - 1468) / (198 / √60)
t = 52 / (198 / √60)
t ≈ 52 / 25.51
t ≈ 2.04
Next, we compare the calculated t-value with the critical t-value from the t-distribution table at the given significance level and degrees of freedom (n - 1).
Since we have a two-tailed test, we divide the significance level (α) by 2 to get the critical values for both tails.
At α = 0.01, the critical t-value for a two-tailed test with 59 degrees of freedom is approximately ±2.66.
Since the calculated t-value (2.04) is less than the critical t-value (2.66), we fail to reject the null hypothesis.
Therefore, there is not sufficient evidence at α = 0.01 to conclude that the health care expenditure of the students differs from the national average of $1468.