-3 < -1, subtract both sides by 3

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

Can someone please help me with this math question?

Ivan

V=(22-2t)(15-2t)(12-2t) and we are told the thickness is 1/2 so

V=(21)14(11)=3234in^3

B :)

6 0

3 years ago

A swimming pool measures 35 feet wide by long. What is the ratio of the length to the perimeter in simpiest form

guapka [62]

<h3>Question:</h3>

A swimming pool measures 35 feet wide by 50 feet long. What is the ratio of the length to the perimeter

<h3>Answer:</h3>

The ratio of length to perimeter of pool is 5 : 17

<h3>Solution:</h3>

Given that, swimming pool measures 35 feet wide by 50 feet long

Therefore,

Length = 50 feet

Width = 35 feet

Swimming pool is usually of rectangular shape

The perimeter of rectangle is given as:

Perimeter = 2(length + width)

Perimeter = 2(50 + 35)

Perimeter = 2(85) = 170

Thus perimeter of pool is 170 feet

Ratio of length to perimeter:

$\frac{Length}{Perimeter } = \frac{50}{170}\\\\\frac{Length}{Perimeter } = \frac{5}{17}$

Thus ratio of length to perimeter of pool is 5 : 17

6 0

3 years ago