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 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
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