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 0.05 alpha level can it be conculded that the variance in the grades differs from 100? Write a 90% confidence level of the variance using the information given

92.3 89.4 76.9 65.2. 49.1. 96 7. 69.5. 72.8. 77.5 52.8. 8.85. 79.2. 72.9. 68.7. 75.8

There are different ways you can do this kind of problem, but the formula below might be one of the easier ways:

s/[1 + (1.645/√2n)]
..to..
s/[1 - (1.645/√2n)]

...where s = standard deviation, 1.645 represents the 90% confidence interval using a z-table, and n = sample size.

You will need to calculate standard deviation from the data given, then substitute the standard deviation and sample size into the formula.

After you finish the above calculations, square the standard deviation values to find the variance values.

I hope this will help get you started.

Well, first of all, let me say that I'm not much of a math whiz, but I'll try my best to answer your question with a touch of humor!

To determine whether the variance in the grades differs from 100, we can perform a statistical test called the chi-square test. The null hypothesis would be that the variance is indeed 100, and the alternative hypothesis would be that it differs from 100.

Now, assuming you've done all the boring calculations and determined the proper test statistic value, you can compare it to the critical chi-square value at the 0.05 alpha level. If the test statistic value exceeds the critical value, then you would reject the null hypothesis and conclude that the variance is different from 100. However, if the test statistic value is smaller than the critical value, you would fail to reject the null hypothesis and conclude that there is not enough evidence to suggest a difference in variance.

Now, moving on to the 90% confidence level estimation of the variance. This means that we can say with 90% confidence that the true variance lies within a certain range. To find this range, we need to calculate the lower and upper bounds of the confidence interval for the variance. Again, I won't bore you with all the mathematical details, but let's just say that it involves some fancy formulas and calculations.

So, to summarize, we need to perform a chi-square test at the 0.05 alpha level to determine if the variance differs from 100. And we can also estimate the 90% confidence interval for the variance. Good luck with all the calculations, and remember, if math gets too difficult, just call in the clowns!

To test if the variance in the grades differs from 100, we can use the Chi-Squared test. Here are the steps to perform the test:

Step 1: Define the null and alternative hypotheses:
- The null hypothesis (H0): The variance in the grades is equal to 100.
- The alternative hypothesis (Ha): The variance in the grades is not equal to 100.

Step 2: Calculate the sample variance:
To find the sample variance, use the formula:
S^2 = Σ(xi - x̄)^2 / (n - 1)
where xi is each grade, x̄ is the mean of the grades, and n is the sample size.

Step 3: Calculate the test statistic:
The test statistic follows a Chi-Squared distribution with (n - 1) degrees of freedom.
Chi-Squared = (n - 1) * S^2 / σ^2
where σ^2 is the hypothetical variance under the null hypothesis (100 in this case).

Step 4: Calculate the critical value:
The critical value for a Chi-Squared test with (n - 1) degrees of freedom at an alpha level of 0.05 is determined using a Chi-Squared distribution table or a statistical software.

Step 5: Compare the test statistic with the critical value:
If the test statistic is greater than the critical value, reject the null hypothesis. Otherwise, fail to reject the null hypothesis.

To calculate the 90% confidence interval for the variance, we can use the Chi-Squared distribution.
The confidence interval is given by:
( (n - 1) * S^2 / χ2α/2, (n - 1) * S^2 / χ2(1 - α/2) )
where χ2α/2 and χ2(1 - α/2) are the Chi-Squared values corresponding to the lower and upper percentile points of the distribution, respectively.

Now, let's calculate the results for the given data:

Step 1: H0: σ^2 = 100 (Variance is equal to 100)
Ha: σ^2 ≠ 100 (Variance is not equal to 100)

Step 2: Calculate the sample variance:
The sample mean (x̄) = (92.3 + 89.4 + 76.9 + 65.2 + 49.1 + 96.7 + 69.5 + 72.8 + 77.5 + 52.8 + 8.85 + 79.2 + 72.9 + 68.7 + 75.8) / 15 ≈ 70.637

Using the variance formula:
S^2 = ( (92.3 - 70.637)^2 + (89.4 - 70.637)^2 + (76.9 - 70.637)^2 + (65.2 - 70.637)^2 + (49.1 - 70.637)^2 + (96.7 - 70.637)^2 + (69.5 - 70.637)^2 + (72.8 - 70.637)^2 + (77.5 - 70.637)^2 + (52.8 - 70.637)^2 + (8.85 - 70.637)^2 + (79.2 - 70.637)^2 + (72.9 - 70.637)^2 + (68.7 - 70.637)^2 + (75.8 - 70.637)^2 ) / 14 ≈ 546.71

Step 3: Calculate the test statistic:
Chi-Squared = (15 - 1) * 546.71 / 100 ≈ 85.394

Step 4: Calculate the critical value:
At an alpha level of 0.05, with (15 - 1) = 14 degrees of freedom, the critical values are χ214/2 = 23.684 and χ2141 - α/2 = 2.146.

Step 5: Compare the test statistic with the critical value:
Since the test statistic (85.394) is greater than the critical value (23.684), we reject the null hypothesis (H0). Therefore, we can conclude that the variance in the grades differs from 100.

The 90% confidence interval for the variance is given by:
((15 - 1) * 546.71 / χ2141 - α/2, (15 - 1) * 546.71 / χ214/2)
((15 - 1) * 546.71 / 2.146, (15 - 1) * 546.71 / 23.684)
(193.35, 44.92)

Therefore, there is a 90% confidence that the true variance lies between 193.35 and 44.92.

To determine if the variance in the grades differs from 100, we can use the Chi-squared test for variance. Here are the steps to calculate and interpret the test:

Step 1: State the hypotheses.
- Null hypothesis (H0): The variance in the grades is equal to 100.
- Alternative hypothesis (Ha): The variance in the grades differs from 100.

Step 2: Set the significance level (alpha).
In this case, the alpha level is given as 0.05.

Step 3: Calculate the test statistic.
To calculate the test statistic, we need to use the following formula:

χ^2 = (n - 1) * s^2 / σ^2

Where:
- χ^2 is the test statistic
- n is the sample size
- s^2 is the sample variance
- σ^2 is the hypothesized population variance (100 in this case)

Using the provided sample of midterm grades, we can calculate the sample variance (s^2) as follows:
s^2 = Σ(x - x̄)^2 / (n - 1)

Given the sample: 92.3, 89.4, 76.9, 65.2, 49.1, 96.7, 69.5, 72.8, 77.5, 52.8, 8.85, 79.2, 72.9, 68.7, 75.8
- Calculate the sample mean (x̄): (92.3 + 89.4 + 76.9 + 65.2 + 49.1 + 96.7 + 69.5 + 72.8 + 77.5 + 52.8 + 8.85 + 79.2 + 72.9 + 68.7 + 75.8) / 15
- Calculate the sum of squared deviations: (92.3 - x̄)^2 + (89.4 - x̄)^2 + (76.9 - x̄)^2 + ... + (75.8 - x̄)^2 / 14
- Divide the sum of squared deviations by (n - 1): sum of squared deviations / 14

This will give us the sample variance (s^2).

Finally, substitute the values into the formula and calculate the test statistic.

Step 4: Determine the critical value.
To determine the critical value, we need to specify the degrees of freedom and the significance level (alpha). The degrees of freedom is the sample size minus 1 (n - 1). Look up the critical value in the Chi-squared distribution table with the specified degrees of freedom and alpha level.

Step 5: Compare the test statistic with the critical value.
If the test statistic is greater than the critical value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.

--------------------------

Additionally, to write a 90% confidence interval for the variance, we can use the Chi-squared distribution. The confidence interval formula is:

[sqrt((n - 1) * s^2 / χ^2_upper), sqrt((n - 1) * s^2 / χ^2_lower)]

Where:
- χ^2_upper and χ^2_lower are the upper and lower critical values from the Chi-squared distribution table, respectively. These critical values correspond to the desired confidence level (90% in this case) and the degrees of freedom (n - 1).

Calculate the square root of the upper and lower bounds of the 90% confidence interval and substitute the values to get the final answer.