What is 4y times 3x.

A new post-surgical treatment is being compared with a standard treatment. Seven subjects receive the new treatment, while seven

AnnyKZ [126]

Answer:

$t=\frac{17.71-28.71}{\sqrt{\frac{4.461^2}{7}+\frac{7.387^2}{7}}}}=-3.372$

$df=n_{S}+n_{N}-2=7+7-2=12$

Since is a one sided test the p value would be:

$p_v =P(t_{(12)}$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean time for the new treatment is significantly lower than the time for the standard treatment at 5% of significance.

Step-by-step explanation:

New: 12, 13,15,19,20,21

Standard: 18,23,24,30,32,35

We can calculate the sample mean and deviation with the following formulas:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

$s= \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

Data given and notation

$\bar X_{N}=17.71$ represent the mean for the new case

$\bar X_{S}=28.71$ represent the mean for the standard case

$s_{N}=4.461$ represent the sample standard deviation for the new case

$s_{S}=7.387$ represent the sample standard deviation for the standard case

$n_{S}=7$ sample size selected for the standard case

$n_{N}=7$ sample size selected for the new case

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if recovery time for patients who receive the new treatment is less than the mean for patients who receive the standard treatment, the system of hypothesis would be:

Null hypothesis: $\mu_{N} \geq \mu_{S}$

Alternative hypothesis: $\mu_{N} < \mu_{S}$

If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{N}-\bar X_{S}}{\sqrt{\frac{s^2_{N}}{n_{N}}+\frac{s^2_{S}}{n_{S}}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{17.71-28.71}{\sqrt{\frac{4.461^2}{7}+\frac{7.387^2}{7}}}}=-3.372$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n_{S}+n_{N}-2=7+7-2=12$

Since is a one sided test the p value would be:

$p_v =P(t_{(12)}$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean time for the new treatment is significantly lower than the time for the standard treatment at 5% of significance.

6 0

4 years ago

I will mark u as brainliest if u answer this!!

lawyer [7]

i)On z, define a∗b=a−b
here aϵz
+
and bϵz
+

i.e.,a and b are positive integers
Let a=2,b=5⇒2∗5=2−5=−3
But −3 is not a positive integer
i.e., −3∈
/
z
+

hence,∗ is not a binary operation.
ii)On Q,define a∗b=ab−1
Check commutative
∗ is commutative if,a∗b=b∗a
a∗b=ab+1;a∗b=ab+1=ab+1
Since a∗b=b∗aforalla,bϵQ
∗ is commutative.
Check associative
∗ is associative if (a∗b)∗c=a∗(b∗c)
(a∗b)∗c=(ab+1)∗c=(ab+1)c+1=abc+c+1
a∗(b∗c)=a∗(bc+1)=a(bc+1)+1=abc+a+1
Since (a∗b)∗c

=a∗(b∗c)
∗ is not an associative binary operation.
iii)On Q,define a∗b=
2
ab

Check commutative
∗ is commutative is a∗b=b∗a
a∗b=
2
ab

b∗a=
2
ba

=
2
ab

a∗b=b∗a∀a,bϵQ
∗ is commutativve.
Check associative
∗ is associative if (a∗b)∗c=a∗(b∗c)
(a∗b)∗c=
2
(
2
ab

)∗c

=
4
abc

(a∗b)∗c=a∗(b∗c)=
2
a×
2
bc

=
4
abc

Since (a∗b)∗c=a∗(b∗c)∀a,b,cϵQ
∗ is an associative binary operation.
iv)On z
+
, define if a∗b=b∗a
a∗b=2
ab

b∗a=2
ba
=2
ab

Since a∗b=b∗a∀a,b,cϵz
+

∗ is commutative.
Check associative.
∗ is associative if $$
(a∗b)∗c=a∗(b∗c)
(a∗b)∗c=(2
ab
)
∗
c=2
2
ab

c
a∗(b∗c)=a∗(2
ab
)=2
a2
bc

Since (a∗b)∗c

=a∗(b∗c)
∗ is not an associative binary operation.
v)On z
+
define a∗b=a
b

a∗b=a
b
,b∗a=b
a

⇒a∗b

=b∗a
∗ is not commutative.
Check associative
∗ is associative if $$
(a∗b)∗c=a∗(b∗c)
(a∗b)∗c=(a
b
)
∗
c=(a
b
)
c

a∗(b∗c)=a∗(2
bc
)=2
a2
bc

eg:−Leta=2,b=3 and c=4
(a∗b)
∗
c=(2∗3)
∗
4=(2
3
)
∗
4=8∗4=8
4

a∗(b∗c)=2
∗
(3∗4)=2
∗
(3
4
)=2∗81=2
81

Since (a∗b)∗c

=a∗(b∗c)
∗ is not an associative binary operation.
vi)On R−{−1}, define a∗b=
b+1
a

Check commutative
∗ is commutative if a∗b=b∗a
a∗b=
b+1
a

b∗a=
a+1
b

Since a∗b

=b∗a
∗ is not commutatie.
Check associative
∗ is associative if (a∗b)∗c=a∗(b∗c)
(a∗b)∗c=(
b+1
a

)
∗
c=
c
b
a

+1

=
c(b+1)
a

a∗(b∗c)=a∗(
c+1
b

)=
c+1
b
a

=
b
a(c+1)

Since (a∗b)∗c

=a∗(b∗c)
∗ is not a associative binary operation

6 0

3 years ago

You created a scale drawing with the scale 1:8. The client asked you to change it to a scale of 1:10. Will your revised drawing

myrzilka [38]

Answer:

The revised drawing be larger or smaller than your original one

Step-by-step explanation:

The revised drawing be larger or smaller than your original one.

Because a drawing at a scale of:

1:10 means that the object is 10 times smaller than in real life scale 1:1

1: 8 means that the object is 8 times smaller than in real life scale 1:1

So it means the revised drawing be larger or smaller than your original one

5 0

3 years ago

Read 2 more answers

Tony purchased a 15-foot- long sub a party.he cut the sandwich into 5/12 foot section into how many pieces dose he cut the sandw

gizmo_the_mogwai [7]

Answer:

Step-by-step explanation:

I really don’t now but you should keep trying!!!

4 0

3 years ago

How do I find y and what is the answer

Citrus2011 [14]

Answer:

y = 25 degrees

Step-by-step explanation:

straight line has 180 degrees

so 180 - 115 = x which is 65 degrees

traingle has 180 also

65 + 90 = 155 180 - 155 = y which is 25 degrees

The triangle on the other side is the equivalent.

6 0

3 years ago

Read 2 more answers