Answer:
The test is not significant at 5% level of significance, hence we conclude that there's no variation among the discussion sections.
Step-by-step explanation:
Assumptions:
1. The sampling from the different discussion sections was independent and random.
2. The populations are normal with means and constant variance
There's no variation among the discussion sections
There's variation among the discussion sections

Df Sum Sq Mean sq F value Pr(>F)
Section 7 525.01 75 1.87 0.99986
Residuals 189 7584.11 40.13
Test Statistic = 

Since our p-value is greater than our level of significance (0.05), we do not reject the null hypothesis and conclude that there's no significant variation among the eight discussion sections.
Answer:
(a) The sample sizes are 6787.
(b) The sample sizes are 6666.
Step-by-step explanation:
(a)
The information provided is:
Confidence level = 98%
MOE = 0.02
n₁ = n₂ = n

Compute the sample sizes as follows:



Thus, the sample sizes are 6787.
(b)
Now it is provided that:

Compute the sample size as follows:

![n=\frac{(z_{\alpha/2})^{2}\times [\hat p_{1}(1-\hat p_{1})+\hat p_{2}(1-\hat p_{2})]}{MOE^{2}}](https://tex.z-dn.net/?f=n%3D%5Cfrac%7B%28z_%7B%5Calpha%2F2%7D%29%5E%7B2%7D%5Ctimes%20%5B%5Chat%20p_%7B1%7D%281-%5Chat%20p_%7B1%7D%29%2B%5Chat%20p_%7B2%7D%281-%5Chat%20p_%7B2%7D%29%5D%7D%7BMOE%5E%7B2%7D%7D)
![=\frac{2.33^{2}\times [0.45(1-0.45)+0.58(1-0.58)]}{0.02^{2}}\\\\=6665.331975\\\\\approx 6666](https://tex.z-dn.net/?f=%3D%5Cfrac%7B2.33%5E%7B2%7D%5Ctimes%20%5B0.45%281-0.45%29%2B0.58%281-0.58%29%5D%7D%7B0.02%5E%7B2%7D%7D%5C%5C%5C%5C%3D6665.331975%5C%5C%5C%5C%5Capprox%206666)
Thus, the sample sizes are 6666.
Use a calculator and change DEG (degres) to RAD (radians).
-60 -49 -40 9
-45 -18 80 82
-74 -23 14 85
-79 -68 21 46
-48 -21 27 54
-68 -13 4 32