HELP PLEASE I'M DESPERATE.

A business school has a goal that the average number of years of work experience of its MBA applicants is more than three years.

gizmo_the_mogwai [7]

Answer:

$z=\frac{3.1-3}{\frac{2.449}{\sqrt{47}}}=0.280$

$p_v =P(z>0.280)=0.390$

If we compare the p value and the significance level assumed $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the height of men actually its NOT significant higher than 0.3 at 1% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=3.1$ represent the sample mean

$\sigma=\sqrt{6}$ represent the sample standard deviation for the sample

$n=47$ sample size

$\mu_o =3$ represent the value that we want to test

$\alpha$ 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 the mean is higher than 3, the system of hypothesis would be:

Null hypothesis: $\mu \leq 3$

Alternative hypothesis: $\mu > 3$

If we analyze the size for the sample is > 30 and we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

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

$z=\frac{3.1-3}{\frac{2.449}{\sqrt{47}}}=0.280$

P-value

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

$p_v =P(z>0.280)=0.390$

Conclusion

If we compare the p value and the significance level assumed $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the height of men actually its NOT significant higher than 0.3 at 1% of signficance.

3 0

2 years ago

The rectangle shown has a perimeter of 48 cm and the given area. Its length is 6 more than twice its width. Write and solve a sy

melamori03 [73]

Answer:

The length of the rectangle is 18 cm

The width of the rectangle is 6 cm

Step-by-step explanation:

Let

x-----> the length of the rectangle

y----> the width of the rectangle

we know that

The perimeter of the rectangle is

$P=2(x+y)$

we have

$P=48\ cm$

so

$48=2(x+y)$ ------> equation A

$x=2y+6$ ------> equation B

Substitute equation B in equation A and solve for y

48=2(2y+6+y)

48=2(3y+6)

48=6y+12

6y=48-12

y=36/6=6 cm

Find the value of x

x=2(6)+6=18 cm

The area of the rectangle is

A=xy

A=18*6

A=108 cm^2

4 0

3 years ago

A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than ex

Olenka [21]

Answer:

a

The 90% confidence interval that estimate the true proportion of students who receive financial aid is

$0.533 < p < 0.64$

b

$n = 1789$

Step-by-step explanation:

Considering question a

From the question we are told that

The sample size is n = 200

The number of student that receives financial aid is $k = 118$

Generally the sample proportion is

$\^ p = \frac{k}{n}$

=> $\^ p = \frac{118}{200}$

=> $\^ p = 0.59$

From the question we are told the confidence level is 90% , hence the level of significance is

$\alpha = (100 - 90 ) \%$

=> $\alpha = 0.10$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.645$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{\frac{\^ p (1- \^ p)}{n} }$

=> $E = 1.645 * \sqrt{\frac{0.59 (1- 0.59)}{200} }$

=> $E = 0.057$

Generally 90% confidence interval is mathematically represented as

$\^ p -E < p < \^ p +E$

=> $0.533 < p < 0.64$

Considering question b

From the question we are told that

The margin of error is E = 0.03

From the question we are told the confidence level is 99% , hence the level of significance is

$\alpha = (100 - 99 ) \%$

=> $\alpha = 0.01$

Generally from the normal distribution table the critical value of is

$Z_{\frac{\alpha }{2} } = 2.58$

Generally the sample size is mathematically represented as

$[\frac{Z_{\frac{\alpha }{2} }}{E} ]^2 * \^ p (1 - \^ p )$

=> $n = [\frac{2.58}{0.03} ]^2 * 0.59 (1 - 0.59 )$

=> $n = 1789$

8 0

2 years ago

If f(x)= x^3-2x+6, find f(-2)

Marina CMI [18]

--8+4+6=2 Hope it helps :-)

5 0

3 years ago

Help me with this question

shutvik [7]

D. 1/6

0.16=1/6

So answers is D

6 0

3 years ago

Read 2 more answers