Mathematics, 21.05.2020 23:04, taterbugwallen7944

A random sample of 15 observations from the first population revealed a sample mean of 350 and a sample standard deviation of 12. A random sample of 17 observations from the second population revealed a sample mean of 342 and a sample standard deviation of 15. At the 0.10 significance level, is there a difference in the population means?

(a) state the decision rule

(b) compute the pooled estimate of the population variance

(c) compute the test statistic

(d) state your decision about the null hypothesis

(e) estimate the p-value.

Answers: 2

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answered: AlexGaming1822

a) df =n_1 +n_2 -2 =15 +17-2 =30

And then since we want to conduct a two tailed test we need to llok in the t distribution with 30 degrees of freedom a value who accumulate 0.05 of the area and we got:

$t_{\alpha/2} =\pm 1.697$

b) $S^2_p =\frac{(15-1)(12)^2 +(17 -1)(15)^2}{15 +17 -2}=187.2$

S_p=13.682

c) $t=\frac{(350 -342)-(0)}{13.682\sqrt{\frac{1}{15}+\frac{1}{17}}}=1.65$

d) For this case the statistic calculated is lower than the critical value so then we FAIL to reject the null hypothesis at the significance level given.

e) The p value for this case would be given by:

$p_v =2*P(t_{30}1.65) =0.1094$

Step-by-step explanation:

We are assuming that the two population deviations are equal:

$\sigma^2_1 =\sigma^2_2 =\sigma^2$

And the statistic is given by this formula:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

Where t follows a t distribution with n_1+n_2 -2 degrees of freedom and the pooled variance S^2_p is given by this formula:

$\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

Part a

For this case we need to find the degrees of freedom given by:

df =n_1 +n_2 -2 =15 +17-2 =30

And then since we want to conduct a two tailed test we need to llok in the t distribution with 30 degrees of freedom a value who accumulate 0.05 of the area and we got:

$t_{\alpha/2} =\pm 1.697$

Part b

The pooled deviation can be founded with this formula:

$S^2_p =\frac{(15-1)(12)^2 +(17 -1)(15)^2}{15 +17 -2}=187.2$

And the deviation would be:

S_p=13.682

Part c

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 = \mu_2$

Alternative hypothesis: $\mu_1 \neq \mu_2$

Replacing the info given we got:

$t=\frac{(350 -342)-(0)}{13.682\sqrt{\frac{1}{15}+\frac{1}{17}}}=1.65$

Part d

For this case the statistic calculated is lower than the critical value so then we FAIL to reject the null hypothesis at the significance level given.

Part e

The p value for this case would be given by:

$p_v =2*P(t_{30}1.65) =0.1094$

answered: Guest

the system of inequity is

$y\leq -x+4$

$3y> x$

step-by-step explanation:

from the given graph it is noticed that we have two related lines. one is solid and another is dashed.

if a line passing through two points, then the equation of line is

$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

the solid line passing through the points (0,4) and (4,0).

$y-4=\frac{0-4}{4-0}(x-0)$

$y=-x+4$

the related equation of solid line is $y=-x+4$ .

similarly, find the related equation of dashed line. the dashed liune passing through the points (0,0) and (3,1).

$y-0=\frac{0-1}{3-0}(x-0)$

$3y=x$

the related equation of dashed line is $3y=x$

from the given graph it is noticed that the point (0,1) lies on the shaded region, therefore the point (0,1) must be satisfied by the related equations.

check the related equations by the point (0,1).

$1=-0+4$

$1=4$

this statement is true if the sign of inequality is < instead of equality. since it is an solid line therefore the sign of inequality will be $\leq$ . the required inequality is

$y\leq -x+4$

similarly, check the second related equation.

$3(1)=0$

it is a dashed line, so this statement is true if the sign of inequality is > . the required inequality is

$3y> x$

15 points write a system of inequalities to represent the shaded portion of the graph

answered: Guest

step-by-step explanation:

12: 00 p.m. to 7: 00 p.m. is a total of 7 hours.

4: 00 p.m. to 7: 00 p.m. is a total of 3 hours.

this means that from 4: 00 p.m to 7: 00 p.m. this represents 3/7 of what was lost from 12: 00 p.m. to 7: 00 p.m.

after knowing this, after doing 11,600 - 6,800 = 4,800

4,800 represents 3/7 of what was lost for 7 hours.

we then divide 4,800 by 3 which gets 1,600.

because we know that 1,600 represents 1/7 of how much water was lost, we just have to multiply by 7, which gets 11,200. after adding 11,200 to how much water was left at 7: 00 p.m. which was 6,800. 11,200 + 6,800 = 18,000

this means that the pool originally had 18,000 gallons in the pool.

you can also check your work by doing 18,000 - 4(1600), which becomes 1800 - 6400 which gets you 11,600.