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.
To test whether the NCL's claim is true, we can set up the following hypotheses:
Null Hypothesis (H₀): The average age of a grandmaster is 55 (μ = 55)
Alternative Hypothesis (H₁): The average age of a grandmaster is not 55 (μ ≠ 55)
To test this, we can use a t-test since we have the sample mean and variance.
The test statistic for a t-test is calculated as:
t = (sample_mean - population_mean) / (sample_standard_deviation / sqrt(sample_size))
First, let's calculate the sample standard deviation:
sample_standard_deviation = sqrt(sample_variance)
= sqrt(168.60)
= 12.99
Now, we can calculate the test statistic:
t = (59.18 - 55) / (12.99 / sqrt(50))
= 4.18 / (12.99 / 7.07)
= 4.18 / 1.839
= 2.27
Next, we need to find the critical value for a 1% level of significance. Since this is a two-tailed test, we divide the significance level by 2 and look up the corresponding value in the t-distribution table. Using a table or statistical software, we find that the critical value for a 1% level of significance with a sample size of 50 is approximately 2.68.
Since the test statistic (2.27) is less than the critical value (2.68), we do not reject the null hypothesis.
Therefore, we do not have enough evidence to conclude that the average age of grandmasters is different from 55 at a 1% level of significance.
To test the claim made by the National Chess League (NCL) that the average age of a grandmaster is 55, we can use a hypothesis test. Let's go through the steps:
Step 1: State the hypothesis.
The null hypothesis (H0): The average age of grandmasters is 55.
The alternative hypothesis (Ha): The average age of grandmasters is not 55.
Step 2: Determine the level of significance.
The given significance level is 1%, which corresponds to a 0.01 level of significance.
Step 3: Analyze the data and calculate the test statistic.
We are given a sample of 50 competitors with a mean age of 59.18 and a variance of 168.60. Since the population variance is unknown, we will use a t-test.
The test statistic (t) for testing the mean can be calculated using the formula:
t = (sample mean - population mean) / (sample standard deviation / √sample size)
In this case:
Sample mean = 59.18
Population mean (claimed by NCL) = 55
Sample standard deviation = √(variance) = √168.60 = 12.99
Sample size = 50
t = (59.18 - 55) / (12.99 / √50)
t = 4.18 / (12.99 / 7.07)
t ≈ 4.18 / 1.84
t ≈ 2.27 (rounded to two decimal places)
Step 4: Determine the critical value.
Since we have a two-tailed test, we need to use the critical value associated with the chosen level of significance (α/2 = 0.01/2 = 0.005).
Consulting a t-distribution table or using statistical software, we find the critical value to be approximately ±2.68 (rounded to two decimal places).
Step 5: Make a decision.
If the absolute value of the calculated t-test statistic is greater than the critical value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
In this case, |2.27| < 2.68, so we fail to reject the null hypothesis.
Step 6: State a conclusion.
Based on the test results, there is not enough evidence to support the claim made by the National Chess League (NCL) that the average age of grandmasters is 55.
To test whether the NCL's claim about the average age of a grandmaster is true, we can use a hypothesis test. Let's set up the null hypothesis and the alternative hypothesis for this test.
Null Hypothesis (H0): The average age of a grandmaster is 55.
Alternative Hypothesis (Ha): The average age of a grandmaster is not 55.
Next, we need to determine the appropriate statistical test to use. Since we have a random sample, know the population variance, and want to compare a sample mean to a hypothesized population mean, we can use a t-test.
The formula for the t-statistic is:
t = (sample mean - hypothesized mean) / (standard deviation / sqrt(sample size))
Given information:
Sample mean (x̄) = 59.18
Hypothesized mean (H0) = 55
Sample standard deviation (s) = sqrt(variance) = sqrt(168.60) = 12.99
Sample size (n) = 50
Plugging in the values:
t = (59.18 - 55) / (12.99 / sqrt(50))
Calculating:
t = 4.18 / (12.99 / 7.07)
t ≈ 4.18 / 1.84
t ≈ 2.27
Next, we need to determine the critical value for a significance level of 1%. Since it is a two-tailed test, we will split the significance level equally into two tails, resulting in a 0.5% in each tail. We can then look up the critical value in the t-distribution table.
The degrees of freedom for this test are (n - 1) = (50 - 1) = 49.
Looking up the critical value for a significance level of 0.5% and 49 degrees of freedom, we find that the critical value is approximately ±2.68.
Since our calculated t-value (2.27) is within the range of -2.68 to 2.68, we fail to reject the null hypothesis. We do not have sufficient evidence to conclude that the average age of a grandmaster is different from 55 based on this sample.
Therefore, at the 1% level of significance, we do not reject the NCL's claim that the average age of a grandmaster is 55.