SOLUTION: A study is done by a community group in two neighboring colleges to determine which graduate students with more maths classes. College A samples 11 graduates. Their average is 4 ma

Question 1185821: A study is done by a community group in two neighboring colleges to determine which graduate students with more maths classes. College A samples 11 graduates. Their average is 4 maths classes with a standard deviation of 1.5 math classes. College B samples 9 grad. Their average is 3.5 maths classes with a standard deviation of 1 maths class. The community group believes that a student who graduate from college A has taken more maths classes, on the average. Both population have a normal distribution. Test at a 1% significance level.
(a) is this a test two means or two proportions?
(b) Are the populations' standard deviation known or unknown?
(c) Which distribution do you use to perform the test?
(d) What is the random variables?
(e) What is the null and alternative hypothesis?
(f) is this test right, left, or two tailed
(g) What is the p-value?
(h) Do you reject or not reject the null hypothesis?
Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
Here's how to address each part of the hypothesis test:
**(a) Two Means or Two Proportions?**
This is a test of *two means* because we are comparing the average number of math classes taken by graduates from two different colleges.
**(b) Known or Unknown Population Standard Deviations?**
The population standard deviations are *unknown*. We are given the *sample* standard deviations.
**(c) Which Distribution to Use?**
Since the population standard deviations are unknown and the sample sizes are small (n1 = 11, n2 = 9), we use the *t-distribution*. Because the sample sizes are small, and the population standard deviations are unknown, we must assume that the population variances are equal.
**(d) What is the Random Variable?**
The random variable is the *difference* between the two sample means: x̄1 - x̄2, where x̄1 is the mean number of math classes for College A graduates and x̄2 is the mean number of math classes for College B graduates.
**(e) Null and Alternative Hypotheses:**
* **Null Hypothesis (H0):** The average number of math classes taken by graduates from College A is equal to or less than the average number of math classes taken by graduates from College B. μ1 ≤ μ2
* **Alternative Hypothesis (H1):** The average number of math classes taken by graduates from College A is greater than the average number of math classes taken by graduates from College B. μ1 > μ2 (This is what the community group believes.)
**(f) Right, Left, or Two-Tailed Test?**
This is a *right-tailed* test because the alternative hypothesis states that the mean for College A is *greater* than the mean for College B.
**(g) What is the p-value?**
1. **Calculate the test statistic:**
Because we are assuming the population variances are equal, we use the following formula for the t-statistic:
```
t = (x̄1 - x̄2) / sqrt(s_p^2 * (1/n1 + 1/n2))
```
Where s_p is the pooled standard deviation, calculated as:
```
s_p = sqrt(((n1-1)*s1^2 + (n2-1)*s2^2) / (n1 + n2 - 2))
```
Plugging in the given values, we get s_p = sqrt(((10)*1.5^2 + (8)*1^2) / 18) = 1.32
Then, the t statistic is calculated as:
```
t = (4 - 3.5) / sqrt(1.32^2 * (1/11 + 1/9)) = 0.5/(1.32*0.447) = 0.84
```
2. **Degrees of Freedom:** df = n1 + n2 - 2 = 11 + 9 - 2 = 18
3. **Find the p-value:** Using a t-table or calculator with df = 18 and t = 0.84, we find the p-value. Since it is a right-tailed test, the p-value is the area to the *right* of the calculated t-statistic. The p-value is approximately 0.208.
**(h) Reject or Not Reject the Null Hypothesis?**
* **Significance Level:** α = 0.01
* **Decision:** Since the p-value (0.208) is *greater* than the significance level (0.01), we *fail to reject* the null hypothesis.
**(Conclusion):**
At a 1% significance level, there is *not* sufficient evidence to conclude that graduates from College A take more math classes on average than graduates from College B.

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