Simplify the expression.

Assume that you have a sample of n 1 equals 6, with the sample mean Upper X overbar 1 equals 50, and a sample standard deviati

tigry1 [53]

Answer:

$t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656$

$df=6+5-2=9$

$p_v =P(t_{9}>2.656) =0.0131$

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

Step-by-step explanation:

Data given

$n_1 =6$ represent the sample size for group 1

$n_2 =5$ represent the sample size for group 2

$\bar X_1 =50$ represent the sample mean for the group 1

$\bar X_2 =38$ represent the sample mean for the group 2

$s_1=7$ represent the sample standard deviation for group 1

$s_2=8$ represent the sample standard deviation for group 2

System of hypothesis

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 \leq \mu_2$

Alternative hypothesis: $\mu_1 > \mu_2$

We are assuming that the population variances for each group are the same

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

The statistic for this case is given by:

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

The pooled variance is:

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

We can find the pooled variance:

$S^2_p =\frac{(6-1)(7)^2 +(5 -1)(8)^2}{6 +5 -2}=55.67$

And the pooled deviation is:

$S_p=7.46$

The statistic is given by:

$t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656$

The degrees of freedom are given by:

$df=6+5-2=9$

The p value is given by:

$p_v =P(t_{9}>2.656) =0.0131$

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

4 0

3 years ago

Please help! I need to figure out how to do this

Nadusha1986 [10]

This is simple, what is the total degree of a triangle? 180
so add up 70 + 30 is 100, the answer to x is 80

7 0

3 years ago

How should I tell my crush I like her? She's my friend and I asked her if she liked me yesterday and she said no but I think she

Helen [10]

Answer:

U just have to make her feel comfortable to trust u that she can tell u that she actually likes u....dont push anything just make her feel appreciated ok?

Step-by-step explanation:

that's all

4 0

3 years ago

You wish to test the claim that p > 33 at a level of significance of a = 0.05 and are given sample 19) statistics n = 5O x =

densk [106]

Answer:

test statistic is 0.176

Step-by-step explanation:

Given Data

p>33

a=0.05

n=50

x=33.3

d(population deviation)=12

Test statistics=?

Solution

Test statistic z=(p-x)\(d/sqrt(50))

z=(33.3-30)\(12\sqrt(50))

z=0.176

5 0

3 years ago

Hey can someone help me? thanks

almond37 [142]

Answer:

Option D.

Step-by-step explanation:

We know that we have a function r(x) that has a horizontal asymptote at y = -8

Then, what can we say about the function:

s(x) = r(x + 2) - 1 ?

Ok, first remember that a horizontal asymptote means that as x, the variable, increases (or decreases), the function eventually tends to a given value, but never actually reaches it.

So as x tends to infinity, we should see an almost horizontal line that tends to y = -8.

So if that happens when x tends to infinity, the same thing will happen when x + 2 tends to infinity, because that "+2" does not add a lot.

Then is easy to conclude that:

r(x + 2) also has a horizontal asymptote at y = -8

And our function is:

s(x) = r(x + 2) - 1

So we are subtracting one, then that horizontal asymptote will be at:

y = -8 - 1 = -9

The function s(x) has a horizontal asymptote at y = -9

And because r(x) is continuous, then:

s(x) = r(x + 2) - 1

is also continuos, as there is nothing added that could change the continuity (we do not have any zero in the denominator or something like that)

Then the correct option is D.

"s(x) is continuous and has a horizontal asymptote at y = -9"

7 0

3 years ago