## Trending News

# I need help in Statistics?

1.company drug test success provides a "1-panel-thc" test for marijuana usage. Among 300 tested subjects, results from 27 subjects were wrong (either a false positive or a false negative) Using a 0.05 Significance Level to test the claim that 10% of results were wrong.

2.The heights are measured for the simple random sample of supermodels Crawford, Bundchen,Pestova, Christenson, Hume, Moss, Campbell, Schiffer, and Taylor. They have a mean height of 700 inches and a standard deviation of 1.5 Inches. Use a 0.01 Significance level to test the claim that the supermodels have heights with a mean that is greater than the mean height of 63.6 inches for the women in the general population.

3. In a study of police gunsire reports during a recent year, it was found that among 540 shots fired by New York City Police, 182 hit their targets and among 283 shots fired by LA police , 77 hit their targets. We want to use a 0.05 Significance Level to test the claim that NYU police and LA police have the same proportion of hits. Test the claim by using a testing hypothesis.

4. In a study of proctored and nonproctored tests in an online intermediate algebra course, researchers obtained data for tet results given below. Using a 0.01 Significance level to test the claim that students taking nonp exams get a higher mean than those taking proctored exams. Test the claim using a hypothesis test.

Group 1(proctored) :n1=30, x̅1= 74.30, s1= 12.87

Group 2 (nonp) :n2=32, x̅= 88.62, s2=22.09

### 1 Answer

- AlanLv 72 months agoFavorite Answer
1st several of these question are reverse of the

normal question given students.

That , they are normally want you to test if things

are different. It is harder to prove that thi,ngs are the same.

However, I used the normal method

to try to prove things are different, because if they are

different, then there definitely not the same.

I am not sure you can really show they are the same.

So possibly the correct answer, you cannot prove they are

the same. So it is a possibility that your textbook, course, or

teacher is a troll asking you impossible quesitons.

This question 1. is a proportion question.

It is called a proportion when your probability is based

on what percentage or proportion of people do something. for a proportional the standard deviation = σ_p = sqrt[ P * ( 1 - P ) / n ]

Unless you actual know the size of the population

and then there is more complicated formula which

add a correction factor.

See this reference.

let's say the NULL hypothesis p = 0.10

the alternate hypothesis p <> 0.10

See this reference

https://stattrek.com/estimation/confidence-interva...

so test this hypothesis ,

we assume p = 10%

and as a decimal = 0.10

so p = 0.10 (as a decimal)

1-p = 1-0.10 = 0.90

n = 300

σ_p = sqrt( 0.9*0.1/ 300)

σ_p = 0.017320508

p_hat (test point) = 27/300 = 0.09

Z_test = (x-mean)/standard deviaton =

Z_test = (0.1-0.09)/ 0.017320508

Z_test = 0.577350269

P(z<0.57) = 0.71566

P(z< 0.58) = 0.71904

I don't know how you show it is the same.

but you don't have enough evidence

to Reject the NULL that they are the same

and cannot support that the claim they are different.

Critical_Z(for a 95 confid. int.) = 1.96

since P(z< 1.96) = 0.975

and a 95 % confidence for a two-tail test

goes from 2.5 % to 97.5 % leaving 95 % in the middle.

since both 0.57 and 0.58 are less than 1.96

you cannot support that is different the 10%

2. For this one, since there are only 9 data points.

You probably cannot use the Central Limit to say

it's normal. You will need to use a t-table.

population mean = 63.6

N = 9 , DF = 8

T_test (for a sample avg) = (x_hat -mean)/(S/sqrt(N))

where S = sample standard deviation

N = number of samples = 9

x_hat = 70

T_test = ( 70-63.5) / (1.5/sqrt(9) ) = 13

so you want x> average

you look for one tailed table value for 0.01 and

DF = 8

T_Critical = 2.896

read from a t-table

since T_Test > T_Critical

13> 2.896

you can support the claim that they have a

higher average height than the standard

population and reject the NULL Hypothesis.

3.

Not sure of this answer, but I followed

Khan Academy examples disregarding my 1st instincts.

so 1st find p and σ_p1 (here this value are for

just samples of the full data, so

call LA test group 1

call NYC test group 2

so find p1 and σ_p1

p1 = 182/540 = 91/270

p1 = approx. 0.337037037

in the standalone case

σ_p1= sqrt ( (91/270)(179/270)/ 540)

σ_p1= approx. 0.020341668

p2 = 77/283

p2= approx. 0.272084806

σ_p2=sqrt( (77/283)*(206/283)/ 283)

σ_p2= 0.02994799

to show the they are the same,

you would like to check if the difference is zero.

D= p2-p1= 0.337037037 - 0.272084806

D = 0.064952231

σ_diff^2 = (σ_p1)^2 + (σ_p2)^2

σ_diff = sqrt( (σ_p1)^2 + (σ_p2)^2 )

My first thought would be to use

individual σ_p1 and σ_p2

but video from Khan academy says

since you are assuming they are equal

then you should use the combined fraction

p1 +p2

p_combined = (77+182)/ (283+540) =

p_combined =0.314702309

so Khan Academy say in video below

https://www.khanacademy.org/math/ap-statistics/two...

since you are assuming they are equal

used p_combined to

the variance for the both o_p1^2 and o_p2^2

so if you follow there advice

use

o_p1 = o_p2= sqrt( 0.314702309 *(1-0.314702309 )/829)

o_p1,2= 0.016187869

(σ_p1)^2= 0.000262047

σ_diff^2 = (σ_p1)^2 + (σ_p2)^2

σ_diff^2= 0.000262047 + 0.000262047

σ_diff= sqrt( 0.000524094)

σ_diff = 0.022893104

(while if you used the original values , you get

0.033 approx.)

so using the Khan Academy assumption

z_test = p_diff/ σ_diff =

z_test = 0.064952231/ 0.022893104 =

z_test = 2.837196376

since this based on the Khan academy assumption

to pool the proportion since you are assume

they are the same

since 2.837 > 1.96 (Z_critical for two-sided 95% check)

you can reject the NULL and support that they

are different and are not the same

However, if you decided to just use the original

individual σ_p1 and σ_p2 ,

then you get the opposite answer

since 1.946 < 1.96

then you cannot support they are different.

4.

Here, I would use Welch t-test equations

https://en.wikipedia.org/wiki/Welch%27s_t-test

Group 1(proctored) :n1=30, x̅1= 74.30, s1= 12.87

Group 2 (nonp) :n2=32, x̅2= 88.62, s2=22.09

1st, you are never going to show that they are

higher , because only test data shows that they are lower.

That is, unless you made a typo.

nonp is 88.62 which is greater than 74.30 so you have no hope

in showing proctor is higher and should have

went home.

so if you agree with wikipedia that a student t-test

with the equations should be used.

s_delta_minus =sqrt(s_1^2/n_1+s_2^2/n_2)

s_delta_minus = sqrt ( 12.87^2/30 + 22.09^2/32 ) =

s_delta_minus= 4.557437122

Wikipedia says use this

Welch–Satterthwaite equation to determine the DF

D.F. = (s_1^2/n_1 + s_2^2/n_2)^2/

((s_1^2/n_1)^2/(n_1-1) + (s_2^2/n_2)^2/(n_2-1).

DF = ( 12.87^2/30 + 22.09^2/32)^2 /

( ((12.87^2)/30)^2/ (29) + ( 22.09^2/32)^2/ 31 ))

DF = 50.44342102

DF = 50 rounded

so we want a one-sided , since the question is .

a greater than

DF = 50 , and alpha =0.01 one sided

t_crtical (DF =50, alpha =0.01) = 2.405

Using this calculator :

https://stattrek.com/online-calculator/t-distribut...

t_test = (x1_bar- x2_bar)/ s_minus_delta

t_test = (74.30 - 88.62) / 4.557437122

t_test = -3.142116856

since -3.14 is not greater than 2.405

You cannot support the claim that the proctored

test score are higher.

If you really wanted to, you could support the claim

that they are lower since -3.14< -2.405 on the low end.

- Login to reply the answers