To test whether the NCL's claim is true, we can use a hypothesis test.
Let's assume the null hypothesis (H0) to be that the average age of grandmasters is 55.
The alternative hypothesis (Ha) would then be that the average age of grandmasters is not equal to 55.
Given that the sample mean (x̄) is 59.18 and the sample variance (s^2) is 168.60, we can calculate the standard deviation (s) as the square root of the variance:
s = √(168.60) ≈ 12.99
Next, we need to calculate the test statistic. In this case, we can use the t-test, since the population standard deviation is unknown.
The formula for the t-test statistic is:
t = (x̄ - μ) / (s / √n)
Where:
x̄ is the sample mean
μ is the population mean (claimed by NCL)
s is the sample standard deviation
n is the sample size
In this case, the sample mean (x̄) is 59.18, the population mean (μ) is 55, the sample standard deviation (s) is 12.99, and the sample size (n) is 50.
t = (59.18 - 55) / (12.99 / √50) ≈ 1.979
Now, we need to determine the critical value for a 1% level of significance. Since this is a two-tailed test, we need to divide the significance level by 2 to get the two critical values.
Using a t-table or a t-distribution calculator with degrees of freedom (df) = n - 1 = 50 - 1 = 49, we find that the critical value for a 1% level of significance is approximately ±2.685.
Since the test statistic (t = 1.979) does not exceed the critical value (±2.685), we fail to reject the null hypothesis (H0).
Therefore, there is not enough evidence to conclude that the average age of grandmasters is different from 55 at a 1% level of significance.
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.
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