The ages of competitors in a grandmaster chess tournament is under dispute. The National Chess League (NCL) claims that the average age of a grandmaster is 55. A random sample of 50 competitors at a recent tournament resulted in a mean of 59,18 and a variance of 168,60. Test at a 1% level of significance whether the NCL's claim is true. The alternative hypothesis is
that the average age of a grandmaster is not 55.
We can use a t-test to test this hypothesis. The formula for the t-test statistic is given by:
t = (sample mean - population mean) / (sample standard deviation / √n)
where n is the sample size.
In this case, the sample mean is 59.18, the population mean is 55, the sample standard deviation is √168.60 = 12.99, and n = 50.
Plugging in the values, we get:
t = (59.18 - 55) / (12.99 / √50) ≈ 2.62
Next, we need to determine the critical value of t for a 1% significance level. Since our alternative hypothesis is that the average age of a grandmaster is not 55, this is a two-tailed test. Therefore, we need to find the t-value that leaves 1% in each tail of the t-distribution with 49 degrees of freedom.
Using a t-distribution table or a statistical software, we find that the critical t-value for a 1% significance level with 49 degrees of freedom is approximately ±2.68.
Since our calculated t-value of 2.62 is less than the critical t-value of 2.68, we fail to reject the null hypothesis. This means that there is not enough evidence to support the NCL's claim that the average age of a grandmaster is 55.
The null hypothesis (H0): The average age of grandmasters is equal to 55.
The alternative hypothesis (H1): The average age of grandmasters is not equal to 55.
To test the NCL's claim, we'll use a t-test since we have a sample mean and variance. Given that the sample size is relatively large (n = 50), we can assume the sampling distribution is approximately normally distributed.
Step 1: Set up the hypotheses:
H0: μ = 55 (population mean is 55)
H1: μ ≠ 55 (population mean is not 55)
Step 2: Determine the significance level (α):
The significance level is given as 1% or 0.01.
Step 3: Calculate the test statistic:
We'll use the t-test formula:
t = (x̄ - μ) / (s / √n)
where x̄ is the sample mean, μ is the population mean (55), s is the sample standard deviation, and n is the sample size.
Given:
Sample mean (x̄) = 59.18
Population mean (μ) = 55
Sample standard deviation (s) =√168.60 ≈ 12.99
Sample size (n) = 50
t = (59.18 - 55) / (12.99 / √50)
t ≈ 4.18
Step 4: Determine the critical value:
Since the alternative hypothesis is two-tailed (μ ≠ 55), we need to find the critical value at α/2 = 0.01/2 = 0.005 significance level. Using a t-distribution table or calculator with 49 degrees of freedom (n - 1), the critical value is approximately ±2.68.
Step 5: Make a decision:
If the absolute value of the test statistic (|t|) is greater than the critical value (2.68), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
|4.18| > 2.68
Therefore, we reject the null hypothesis.
Step 6: Conclusion:
Based on the sample data, there is enough evidence to suggest that the average age of grandmasters is not 55.
To test the NCL's claim, we can perform a hypothesis test. The null hypothesis, denoted as H0, represents the claim being tested, while the alternative hypothesis, denoted as Ha, represents the claim opposing the null hypothesis.
In this case, the NCL's claim is that the average age of a grandmaster is 55. Therefore, the null hypothesis is:
H0: The average age of a grandmaster is 55.
The alternative hypothesis opposes the null hypothesis and could be one-sided or two-sided. Since the question does not specify the type of alternative hypothesis, we will assume it to be a two-sided alternative hypothesis:
Ha: The average age of a grandmaster is not equal to 55.
To test this hypothesis, we can use a t-test because we have a sample mean and a known population variance.
First, we need to calculate the test statistic, which is t. The formula for calculating t is:
t = (sample mean - hypothesized mean) / (standard deviation / sqrt(sample size))
Given:
Sample mean (x̄) = 59.18
Hypothesized mean (μ) = 55
Sample size (n) = 50
Sample variance (s^2) = 168.60
To calculate the standard deviation (s), we take the square root of the sample variance:
s = sqrt(sample variance) = sqrt(168.60) ≈ 12.99
Now, we can calculate the test statistic:
t = (59.18 - 55) / (12.99 / sqrt(50))
After calculating t, we will compare it to the critical value from the t-distribution table at a 1% significance level (α = 0.01). The degrees of freedom for a t-test with a known variance is (n - 1).
If the calculated t-value is greater than or less than the critical value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
Note: Since the alternative hypothesis is two-sided, we need to compare the absolute value of the calculated t-value with the critical value.
Once you have the t-value and the critical value, you can determine whether to reject the null hypothesis or fail to reject it based on the comparison.