Question

1.

I am interested in if there is a pay difference between men and women nurses. Based on a study by ucsf male nurses used to make more.  I randomly sample and collect data. Assume you can pool the variances. What is the p-value based on the following sample...

Females Males

xbar 44,000 47,000

s 12,000 12,600   

n 60 57

What is the p-value?

2.

I am interested in if there is a pay difference between men and women nurses. Based on a study by ucsf male nurses make more. To study this I send an email to collect data. In my email I detail the study and ask confidentially for their income. I send the email to registered nurses all over the country based on hospitals who publish emails for nurses. I get a 20% response rate. Which of the following could be a bias of my results? Mark all that apply...

Group of answer choices

Men might possibly make more than women

Only certain hospitals have their nurses emails public facing

Only men who feel they make enough respond

Only women who feel they should be making more reply

Only women who feel they aren't making enough respond

Only men who feel they are making an acceptable wage repy.

for nurses, men make more then women

hospitals that publish emails for nurses may be more likely to pay evenly.

4.

I want to know if people who love turkey eat more calories during thanksgiving compared to people who don't eat turkey. Which type off test could I do?

Group of answer choices

2 proportion Z test

2 Sample Z-test

Chi-Squared Independence

2 Sample T-Test

2 Proportion Confidence Interval

5.

This is real data on pregnant mothers in the hospitals at a certain time. Conduct a hypothesis test to see if people are more likely to smoke with subsequent children than with a first child.  

	Non Smoker	Smoker
Not first child	548	363
First Child	194	121

Answer 1

1)

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 ╪   0                  
                          
Level of Significance ,    α =    0.1                  
                          
Sample #1   ---->   1                  
mean of sample 1,    x̅1=   44000.000                  
standard deviation of sample 1,   s1 =    12000.0000                  
size of sample 1,    n1=   60                  
                          
Sample #2   ---->   2                  
mean of sample 2,    x̅2=   47000.000                  
standard deviation of sample 2,   s2 =    12600.0000                  
size of sample 2,    n2=   57                  
                          
difference in sample means =    x̅1-x̅2 =    44000.0000   -   47000.0   =   -3000.00  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    12295.8317                  
std error , SE =    Sp*√(1/n1+1/n2) =    2274.2478                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -3000.0000   -   0   ) /    2274.25   =   -1.319
                          
Degree of freedom, DF=   n1+n2-2 =    115                  
  
p-value =        0.1898   (excel function: =T.DIST.2T(t stat,df) )              

5)

Ho:   p1 - p2 =   0          
Ha:   p1 - p2 >   0          
                  
sample #1   ----->              
first sample size,     n1=   911          
number of successes, sample 1 =     x1=   363          
proportion success of sample 1 , p̂1=   x1/n1=   0.3985          
                  
sample #2   ----->              
second sample size,     n2 =    315          
number of successes, sample 2 =     x2 =    121          
proportion success of sample 1 , p̂ 2=   x2/n2 =    0.3841          
                  
difference in sample proportions, p̂1 - p̂2 =     0.3985   -   0.3841   =   0.0143
                  
pooled proportion , p =   (x1+x2)/(n1+n2)=   0.3948          
                  
std error ,SE =    =SQRT(p*(1-p)*(1/n1+ 1/n2)=   0.03195          
Z-statistic = (p̂1 - p̂2)/SE = (   0.014   /   0.0319   ) =   0.4487
                  

p-value =        0.3268   [excel function =NORMSDIST(-z)]      
decision :    p-value>α,Don't reject null hypothesis