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Do political science classes require less writing than history classes? The 55 randomly selected political science classes assig

saul85 [17]

Answer:

We conclude that political science classes require less writing time than history classes at 0.01 level of significance.

Step-by-step explanation:

We are given that the 55 randomly selected political science classes assigned an average of 16.3 pages of essay writing for the course. The standard deviation for these 55 classes was 4.8 pages.

The 54 randomly selected history classes assigned an average of 18.9 pages of essay writing for the course. The standard deviation for these 54 classes was 5 pages.

Let $\mu_1$ = average writing time required by political science classes.

$\mu_2$ = average writing time required by history classes.

So, Null Hypothesis, $H_0$ : $\mu_1\geq\mu_2$ {means that political science classes require more or equal writing time than history classes}

Alternate Hypothesis, $H_A$ : $\mu_1$ {means that political science classes require less writing time than history classes}

The test statistics that would be used here Two-sample t test statistics as we don't know about the population standard deviation;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t__n_1_-_n_2_-_2$

where, $\bar X_1$ = sample average pages of essay writing for political science classes = 16.3 pages

$\bar X_2$ = sample average pages of essay writing for history classes = 18.9 pages

$s_1$ = sample standard deviation for political science classes = 4.8 pages

$s_2$ = sample standard deviation for history classes = 5 pages

$n_1$ = sample of political science classes = 55

$n_2$ = sample of history classes = 54

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(55-1)\times 4.8^{2}+(54-1)\times 5^{2} }{55+54-2} }$ = 4.90

So, test statistics = $\frac{(16.3-18.9)-(0)}{4.9 \sqrt{\frac{1}{55}+\frac{1}{54} } }$ ~ $t_1_0_7$

= -2.77

The value of t test statistics is -2.77.

Now, at 0.01 significance level the t table gives critical value of -2.365 at 107 degree of freedom for left-tailed test.

Since our test statistics is less than the critical value of t, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that political science classes require less writing time than history classes.

3 0

3 years ago

Evaluate 47-5 times 3

Vlad1618 [11]

Using order of operations you would multiply first giving you 15. So 47-15=32.

4 0

3 years ago

ASAP PLEASE HELP Your test scores in one class are 82 and 88. What possible scores can you earn on your next test to have a test

Rom4ik [11]

Answer:

85 to 100

Step-by-step explanation:

5 0

4 years ago

The accompanying Venn diagram illustrates a sample space containing six sample points and three events, A, B, and C. The probabi

Ede4ka [16]

Answer:

Step-by-step explanation:

To find the probability of a certain set, we must add the probability of the points that are contained in the specific set.

a) P(A): Add the probability of 1,3,5. That is P(1)+P(3)+P(5) = 0.5

b) $P(A\cap B)$ . This set has no points in it, so its probability is 0.

c) P(A\cup B \cup C): Add the probability of 1,2,3,4,5,6. So, it is 1.

d) $P(C^c) = 1- P(C)$ . P(C) is the probability by summing 5,6,2 so P(C) = 0.5. So $P(C^c)=0.5$

e) $P(A\cap C^c)$ That is, all the points that are in A but not in C. So add 1,3. Then the probability is 0.4

e') $P(B|A) = \frac{P(A\cap B )}{P(A)}=\frac{0}{0.5}=0$ , using the definition of conditional probabily and results a,b.

f) Two events are mutually exlusive if the probability of their intersection is 0. In this case A and C are not mutually exclusive, since $P(A\cap C)$ is the probability of 5, that is 0.1. Since 0.1>0, they are not mutually exclusive.