1) In a sample of 45 women who worked at company their average yearly salary was $63,000 with a standard deviation of $10,000. In a sample of 56 men at the company in similar jobs their average yearly salary was $67,000 with a standard deviation of $12,500.

a) At the .01 level is there a difference in the average salary of the men and women at this company? (p = .084)

 

b) Based on the answer for a) would a claim by a woman of sexism in salaries be legitimate? Explain

 

 2. A sample of 100 people were given a tour of a winery and then shown the process of how the wine was produced. The group of 100 was then given samples to taste of the wine and 48 rated the wine as "excellent." For a separate group of 120 people who were simply given the wine without the tour or being shown the process 20 rated the wine as "excellent" a) At the .05 level was there a difference in the % of people who rated the wine excellent? (p = 0)

b) Based on the answer for a) what appears to be the reason for the difference in the % of the two groups?

 

6. Currently a company has 47% of their employees enrolled in a 401(k) program (this is a retirement program with tax deferrals on income). The company wishes to increase the % of employees in this program. To do that they require each employee to attend a seminar that demonstrates the incredible savings impact that the 401(k) offers for the employees. In a sample of 120 employees who attended the seminar 89 joined the 401(k) program.

a) At the .01 level did the seminar increase the % of employees in the program? (p-value = 0)

b) Interpret the p-value obtained in a)

c) Find and interpret the 99% confidence interval (computer says "lower bound is .6474")

 

 

 3. You sell big screen TV's for a price of $780. You wish to see if the consumer's idea of the price of your TVs is different. In a sample of 27 consumers, when asked "how much do you think this big screen TV costs?" the average price was $830 with a standard deviation of $150.

 

a) At the .01 level is the consumer's idea of the average price different from your price? (p = .095)

 

b) Find and interpret the 99% confidence interval (computer gives (749.79, 910.21))

 

c) If I take another sample of 27 consumers and ask them how much they think the big screen TV costs would an average price of $800 be something that you would expect to occur? explain

 

d) What is the only type of error that could have occurred in this problem? (no explanation of error needed)

 

 4. Engagement rings on average cost $6000. In a sample of 56 dating couples they were asked how much they would spend on their engagement rings and the average was $5300 with a standard deviation of $1200.

 

a) At the .05 level is the price of what the couples were willing to spend statistically significantly different than average? (p = 0)

 

b) Find and interpret the 95% confidence interval (computer says (4978.6,5621.4)) (11)

 

5. In a roulette wheel the color red occurs 48.7% of the time. You think the casino is cheating because the number red appears to be coming up less than expected. Why do you say this? In  a sample of the last 230 spins of the roulette wheel red came up 105 times.

 

a) At the .01 level is red coming up at a statistically significant lower amount than expected? (p-value = .178)

 

b) What is the reason for the difference for red coming up at a lower % based on the answer for a)

(Q1)

(a)Ho: u1 - u2 =0

H1: u1 - u2 ≠ 0

test statistic: -1.7866

critical value: -2.6264 , 2.6264

p-value: 0.077

fail to reject Ho

(b) there is a difference in the average salary of the men and women at this company

----------------------------------------------------------------------------------------

(Q2)

(a)

null, Ho: p1 = p2 

alternate, H1: p1 ≠ p2

test statistic: 5.0076

critical value: -1.96 , 1.96

decision: reject Ho

p-value: 0

(b) there is a difference in the % of people who rated the wine excellent

----------------------------------------------------------------------------------------

(Q6)

(a)

null, Ho:p=0.47 

alternate, H1: p>0.47

test statistic: 5.9627

critical value: 2.326

decision: reject Ho

p-value: 0

evidence that seminar increase the % of employees in the program

(b)

p-value is the probability of rejecting the null hypothesis when in fact the statement is true

(c) 99% confidence interval [0.6387 , 0.8446] ;

interpretations:

1. we are 99% sure that the interval [ 0.6387 , 0.8446] contains the true population proportion

2. if a large number of samples are collected, and a confidence interval is created

----------------------------------------------------------------------------------------

(Q3)

(a)

null, Ho: μ=780 

alternate, H1: μ≠780

test statistic: 1.732

critical value: -2.7787 , 2.7787

decision: fail to reject Ho

p-value: 0.0951

consumer's idea of the average price is not different from your price

(b)

99% confidence level is [ 749.777 , 910.223 ]

(c)

yes, since the value $800 lies in the interval calculates above

(d) 

we see type II error occur in this context

----------------------------------------------------------------------------------------

(Q4)

(a)

null, Ho: μ=6000 

alternate, H1: μ≠6000

test statistic: -4.365

critical value: -2.004 , 2.004

decision: reject Ho

p-value: 0.0001

couples were willing to spend statistically significantly different than average

(b) 95% confidence interval is [ 4978.645 , 5621.355 ]

----------------------------------------------------------------------------------------

(Q5)

(a)

null, Ho:p=0.487 

alternate, H1: p<0.487

test statistic: -0.9248

critical value: -2.326

decision: fail to reject Ho

p-value: 0.17754

At the .01 no evidence to support the fact that it lower amount than expected

(b) there is good chance that it is cheated intentionally to not to appear red

Step-by-step explanation

(Q1)

Given that,

mean(x)=63000

standard deviation , s.d1=10000

number(n1)=45

mean(y)=67000

standard deviation, s.d2 =12500

number(n2)=56

null, Ho: u1 - u2 =0

alternate, H1: u1 - u2 ≠ 0

we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)

to =63000-67000/sqrt((100000000/45)+(156250000/56))

to =-1.7866

| to | =1.7866

critical value

level of significance, alpha = 0.01

degrees of freedom(df) = ( sd1 ^2 / n1 + sd2 ^2 /n2 )^2 / (s1^4 / n1^2 ( n1-1)) + (s2^4 / n2^2 ( n2-1))

df = (( (10000^2/45)+(12500^2/56) ))^2 / (( 10000^4 / (45^2 ( 45 - 1 )) ) + (12500^4/(56^2(56-1))))

df = 98.9995 ~99

from standard normal table, two tailed t alpha/2 =2.6264

since our test is two-tailed,

reject Ho, if to < -2.6264 OR if to > 2.6264

we got |to| = 1.78664 & | t alpha | = 2.6264

make decision

hence value of |to | < | t alpha | and here we fail to reject Ho

p-value: two tailed ( double the one tail ) - Ha : ( p ≠ -1.7866 ) = 0.077

hence value of p0.01 < 0.077,here we fail to reject Ho

----------------------------------------------------------------------------------------------------------------------------------

(Q2)

Given that,

sample one, x1 =48, n1 =100, p1= x1/n1=0.48

sample two, x2 =20, n2 =120, p2= x2/n2=0.1667

finding a p^ value for proportion p^=(x1 + x2 ) / (n1+n2) 

p^=0.3091

q^ Value For Proportion= 1-p^=0.6909

null, Ho: p1 = p2 

alternate, H1: p1 ≠ p2

level of significance, α = 0.05

from standard normal table, two tailed z α/2 =1.96

since our test is two-tailed

reject Ho, if zo < -1.96 OR if zo > 1.96

we use test statistic (z) =  (p1-p2)/√(p^q^(1/n1+1/n2))

zo =(0.48-0.1667)/sqrt((0.309*0.6909(1/100+1/120))

zo =5.0076

| zo | =5.0076

critical value

the value of |z α| at los 0.05% is 1.96

we got |zo| =5.008 & | z α | =1.96

make decision

hence value of | zo | > | z α| and here we reject Ho

p-value: two tailed ( double the one tail ) - Ha : ( p ≠ 5.0076 ) = 0

hence value of p0.05 > 0,here we reject Ho

----------------------------------------------------------------------------------------

(Q6)

(a) Given that,

possible chances (x)=89

sample size(n)=120

success rate ( p )= x/n = 0.7417

success probability,( po )=0.47

failure probability,( qo) = 0.53

null, Ho:p=0.47  

alternate, H1: p>0.47

level of significance, alpha = 0.01

from standard normal table,right tailed z alpha/2 =2.326

since our test is right-tailed

reject Ho, if zo > 2.326

we use test statistic z proportion = p-po/sqrt(poqo/n)

zo=0.74167-0.47/(sqrt(0.2491)/120)

zo =5.9627

| zo | =5.9627

critical value

the value of |z alpha| at los 0.01% is 2.326

we got |zo| =5.963 & | z alpha | =2.326

make decision

hence value of | zo | > | z alpha| and here we reject Ho

p-value: right tail - Ha : ( p > 5.96266 ) = 0

hence value of p0.01 > 0,here we reject Ho

-------------------------------------------------------------------------------------------------------

(c)

success rate ( p )= x/n = 0.7417

level of significance, α = 0.01

from standard normal table, two tailed z α/2 =2.576

CI  = confidence interval 

confidence interval = [ 0.7417 ± 2.576 * sqrt ( (0.7417*0.2583) /120) ) ]

= [0.7417 - 2.576 * sqrt ( (0.7417*0.2583) /120) , 0.7417 + 2.576 * sqrt ( (0.7417*0.2583) /120) ]

= [0.6387 , 0.8446] 

-----------------------------------------------------------------------------------------------

(Q3)

(a)

Given that,

population mean(u)=780

sample mean, x =830

standard deviation, s =150

number (n)=27

null, Ho: μ=780 

alternate, H1: μ≠780

level of significance, alpha = 0.01

from standard normal table, two tailed t alpha/2 =2.7787

since our test is two-tailed

reject Ho, if to < -2.7787 OR if to > 2.7787

we use test statistic (t) = x-u/(s.d/sqrt(n))

to =830-780/(150/sqrt(27))

to =1.732

| to | =1.732

critical value

the value of |t alpha| with n-1 = 26 d.f is 2.7787

we got |to| =1.732 & | t alpha | =2.7787

make decision

hence value of |to | < | t alpha | and here we fail to reject Ho

p-value :two tailed ( double the one tail ) - Ha : ( p ≠ 1.7321 ) = 0.0951

hence value of p0.01 < 0.0951,here we fail to reject Ho

-------------------------------------------------------------------------------------------

(b)

from standard normal table, two tailed value of |t alpha/2| with n-1 = 26 d.f is 2.779

we use CI = x ± t a/2 * (sd/ sqrt(n))

where, 

x  = mean

sd  = standard deviation

a  = 1 - (confidence level/100)

ta/2 = t-table value

CI  = confidence interval 

confidence interval = [ 830 ± t a/2 ( 150/ sqrt ( 27) ] 

= [ 830-(2.779 * 28.868) , 830+(2.779 * 28.868) ]

= [ 749.777 , 910.223 ]

-----------------------------------------------------------------------------------------------

(Q4)

Given that,

population mean(u)=6000

sample mean, x =5300

standard deviation, s =1200

number (n)=56

null, Ho: μ=6000 

alternate, H1: μ≠6000

level of significance, alpha = 0.05

from standard normal table, two tailed t alpha/2 =2.004

since our test is two-tailed

reject Ho, if to < -2.004 OR if to > 2.004

we use test statistic (t) = x-u/(s.d/sqrt(n))

to =5300-6000/(1200/sqrt(56))

to =-4.365

| to | =4.365

critical value

the value of |t alpha| with n-1 = 55 d.f is 2.004

we got |to| =4.365 & | t alpha | =2.004

make decision

hence value of | to | > | t alpha| and here we reject Ho

p-value :two tailed ( double the one tail ) - Ha : ( p ≠ -4.3653 ) = 0.0001

hence value of p0.05 > 0.0001,here we reject Ho

-------------------------------------------------------------------------------------------

(b)

from standard normal table, two tailed value of |t alpha/2| with n-1 = 55 d.f is 2.004

we use CI = x ± t a/2 * (sd/ sqrt(n))

where, 

x  = mean

sd  = standard deviation

a  = 1 - (confidence level/100)

ta/2 = t-table value

CI  = confidence interval 

confidence interval = [ 5300 ± t a/2 ( 1200/ sqrt ( 56) ] 

= [ 5300-(2.004 * 160.357) , 5300+(2.004 * 160.357) ]

= [ 4978.645 , 5621.355 ]

----------------------------------------------------------------------------------------

(Q5)

(a)

Given that,

possible chances (x)=105

sample size(n)=230

success rate ( p )= x/n = 0.4565

success probability,( po )=0.487

failure probability,( qo) = 0.513

null, Ho:p=0.487  

alternate, H1: p<0.487

level of significance, alpha = 0.01

from standard normal table,left tailed z alpha/2 =2.326

since our test is left-tailed

reject Ho, if zo < -2.326

we use test statistic z proportion = p-po/sqrt(poqo/n)

zo=0.45652-0.487/(sqrt(0.249831)/230)

zo =-0.9248

| zo | =0.9248

critical value

the value of |z alpha| at los 0.01% is 2.326

we got |zo| =0.925 & | z alpha | =2.326

make decision

hence value of |zo | < | z alpha | and here we fail to reject Ho

p-value: left tail - Ha : ( p < -0.92476 ) = 0.17754

hence value of p0.01 < 0.17754,here we fail to reject Ho

-------------------------------------------------------------------------------------------------------