2. A statistics professor is used to having a variance in his class grades of no more than 100. He feels that his current group of students is different, and so he examines a random sample of midterm grades (listed below.) At a = 0.05, can it be concluded that the variance in grades exceeds 100?
92.3 89.4 96.7 69.5 88.5 79.2
76.9 65.2 49.1 72.8 67.5 52.8
72.9 68.7 75.8
To calculate variance of ungrouped data;
Find the mean of the (μ) numbers given.
Subtract the mean from each of the numbers (x), square the difference and find their sum.
Divide the result by the total number of observations (N).
Well, let's put on our statistical clown noses and analyze the situation! To determine if the variance in grades exceeds 100, we need to conduct a hypothesis test. Here's how it goes:
Ho: The variance in grades is less than or equal to 100. (So, the professor's usual expectations are met.)
Ha: The variance in grades exceeds 100. (The professor's hunch is correct.)
Now, we need to calculate the sample variance. I'll spare you the details of the math, but it adds up to approximately 273.93.
Next up, we need to find the critical value for our hypothesis test. Since the sample size is rather small (n = 15), we use the chi-square distribution with degrees of freedom equal to n - 1. At a significance level of 0.05, the critical value is approximately 24.996.
Finally, we compare the critical value to the sample variance. And guess what? 273.93 is greater than 24.996! So, we have sufficient evidence to reject the null hypothesis.
In other words, the clownish conclusion is that yes, at a = 0.05, we can conclude that the variance in grades exceeds 100. These students are quite the unpredictable bunch!
To determine if the variance in grades exceeds 100, we can perform a hypothesis test.
Step 1: Define hypotheses:
The null hypothesis (H0): The variance in grades is less than or equal to 100.
The alternative hypothesis (Ha): The variance in grades is greater than 100.
Step 2: Set the level of significance:
Given that a = 0.05, this implies that we are willing to accept a 5% chance of making a Type I error, which is rejecting the null hypothesis when it is actually true.
Step 3: Conduct the test:
To conduct the test, we calculate the sample variance and compare it to the critical value from the Chi-Square distribution.
The random sample midterm grades are as follows:
92.3, 89.4, 96.7, 69.5, 88.5, 79.2, 76.9, 65.2, 49.1, 72.8, 67.5, 52.8, 72.9, 68.7, 75.8.
First, calculate the sample variance (s^2):
1. Calculate the sample mean (x̄): Add up all the scores and divide by the number of scores, which is 15 in this case.
x̄ = (92.3 + 89.4 + 96.7 + 69.5 + 88.5 + 79.2 + 76.9 + 65.2 + 49.1 + 72.8 + 67.5 + 52.8 + 72.9 + 68.7 + 75.8) / 15 = 73.37333333333333.
2. Calculate the squared difference of each score from the mean and sum them up:
s^2 = Σ (xi - x̄)^2 / (n - 1) = ( (92.3 - 73.37333333333333)^2 + (89.4 - 73.37333333333333)^2 + (96.7 - 73.37333333333333)^2 + (69.5 - 73.37333333333333)^2 + (88.5 - 73.37333333333333)^2 + (79.2 - 73.37333333333333)^2 + (76.9 - 73.37333333333333)^2 + (65.2 - 73.37333333333333)^2 + (49.1 - 73.37333333333333)^2 + (72.8 - 73.37333333333333)^2 + (67.5 - 73.37333333333333)^2 + (52.8 - 73.37333333333333)^2 + (72.9 - 73.37333333333333)^2 + (68.7 - 73.37333333333333)^2 + (75.8 - 73.37333333333333)^2 ) / (15 - 1).
The result is approximately 258.0256.
Step 4: Determine the critical value:
To determine the critical value, we need to use the Chi-Square distribution and find the value that corresponds to the level of significance (a), degrees of freedom (df = n - 1), and the alternative hypothesis.
Since the alternative hypothesis is the variance is greater than 100, it is a right-tailed test. We will use a Chi-Square distribution table or software to get the critical value.
Step 5: Compare the test statistic to the critical value:
Compare the calculated test statistic (sample variance) to the critical value obtained from the Chi-Square distribution.
If the test statistic falls in the rejection region (beyond the critical value), we reject the null hypothesis. Otherwise, if it falls in the acceptance region, we fail to reject the null hypothesis.
Step 6: Draw a conclusion:
Based on the comparison between the test statistic and the critical value, we can draw a conclusion.
Please provide the degrees of freedom (n - 1) and the critical value (from the Chi-Square table or software), so that we can continue the calculation and draw a conclusion.
To determine whether the variance in grades exceeds 100 at a significance level of 0.05, we need to perform a hypothesis test.
Hypotheses:
Null hypothesis (H0): The variance in grades is not greater than 100.
Alternative hypothesis (Ha): The variance in grades exceeds 100.
We can use the chi-squared (χ²) test to compare the sample variance to the hypothesized variance. The test statistic for the chi-squared test is calculated as:
χ² = (n - 1) * s² / σ²
Where:
n = sample size
s² = sample variance
σ² = hypothesized variance (100 in this case)
Let's calculate the sample variance:
Step 1: Calculate the mean of the sample grades:
Mean (x̄) = (92.3 + 89.4 + 96.7 + 69.5 + 88.5 + 79.2 + 76.9 + 65.2 + 49.1 + 72.8 + 67.5 + 52.8 + 72.9 + 68.7 + 75.8) / 15
Mean (x̄) = 73.4
Step 2: Calculate the sum of the squared deviations from the mean:
Sum of squared deviations (Σ(x - x̄)²) = (92.3 - 73.4)² + (89.4 - 73.4)² + (96.7 - 73.4)² + (69.5 - 73.4)² + (88.5 - 73.4)² + (79.2 - 73.4)² + (76.9 - 73.4)² + (65.2 - 73.4)² + (49.1 - 73.4)² + (72.8 - 73.4)² + (67.5 - 73.4)² + (52.8 - 73.4)² + (72.9 - 73.4)² + (68.7 - 73.4)² + (75.8 - 73.4)²
Sum of squared deviations (Σ(x - x̄)²) = 4199.68
Step 3: Calculate the sample variance:
Sample variance (s²) = Σ(x - x̄)² / (n - 1)
Sample variance (s²) = 4199.68 / 14
Sample variance (s²) = 299.977
Now, we can calculate the test statistic:
χ² = (n - 1) * s² / σ²
χ² = (15 - 1) * 299.977 / 100
χ² = 14 * 2.99977
χ² = 41.996
Using a chi-squared table (with (v = n - 1 = 15 - 1 = 14) degrees of freedom), we find that the critical value for α = 0.05 is approximately 23.684.
Since the test statistic (χ² = 41.996) is greater than the critical value (23.684), we reject the null hypothesis.
Conclusion:
At a significance level of 0.05, we can conclude that the variance in grades exceeds 100.