do political science classes require the same...
do political science classes require the same amount of writing as history classes? the 49 randomly selected political science classes assigned an average of 15.5 pages of essay writing for the course. the standard deviation for these 49 classes was 5.9 pages. the 46 randomly selected history classes assigned an average of 19.1 pages of essay writing for the course. the standard deviation for these 46 classes was 6 pages. what can be concluded at the α = 0.05 level of significance? for this study, we should use select an answer a. the null and alternative hypothese select an answer (h_0:mu1) (h_1:mu1) b. the test statistic? = -2.946 c. the p - value = 0.0032 (please d. the p - value is s α e. based on this, we should reject f. thus, the final conclusion is that. the results are statistically conclude that the populatior require is not the same as th require. the results are statistically evidence to conclude that th science classes require is eq classes require. the results are statistically that the population mean not the same as th
Answer
# Explanation:
## Step1: Identify test type
We are comparing the means of two independent groups (political - science classes and history classes), and the population standard - deviations are unknown. So, we use a t - test for the difference between two independent population means.
## Step2: State hypotheses
Let $\mu_1$ be the population mean of pages of essay writing for political science classes and $\mu_2$ be the population mean of pages of essay writing for history classes.
$H_0:\mu_1=\mu_2$
$H_1:\mu_1\neq\mu_2$
## Step3: Calculate test statistic
The formula for the t - test statistic for two independent samples is $t=\frac{(\bar{x}_1 - \bar{x}_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$, where $\bar{x}_1 = 15.5$, $n_1 = 49$, $s_1 = 5.9$, $\bar{x}_2=19.1$, $n_2 = 46$, $s_2 = 6$. First, we calculate the pooled standard deviation $s_p=\sqrt{\frac{(n_1 - 1)s_1^2+(n_2 - 1)s_2^2}{n_1 + n_2-2}}$. After substituting the values, we get $s_p=\sqrt{\frac{(49 - 1)\times5.9^2+(46 - 1)\times6^2}{49+46 - 2}}\approx5.95$. Then $t=\frac{(15.5 - 19.1)-0}{5.95\sqrt{\frac{1}{49}+\frac{1}{46}}}\approx - 2.946$.
## Step4: Calculate p - value
The degrees of freedom is $df=n_1 + n_2-2=49 + 46-2 = 93$. Using a t - distribution table or software, for a two - tailed test with $t=-2.946$ and $df = 93$, the p - value is approximately $0.0032$.
## Step5: Make a decision
Since the p - value ($0.0032$) is less than $\alpha = 0.05$, we reject the null hypothesis.
## Step6: State conclusion
The results are statistically significant. There is sufficient evidence to conclude that the population mean number of pages of writing required for political science classes is not the same as that for history classes.
# Answer:
a. $H_0:\mu_1=\mu_2$, $H_1:\mu_1\neq\mu_2$
b. $t=-2.946$
c. $p - value = 0.0032$
d. $p - value<\alpha$
e. Reject $H_0$
f. The results are statistically significant. There is sufficient evidence to conclude that the population mean number of pages of writing required for political science classes is not the same as that for history classes.