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it is an 8 question test that I have uploaded from a scanned document. if there are any questions please call

David · Accepted Answer

For all the questions consider µ1 and µ2 are population means,   1
2
 and  2
2
 are the population 
variances, ̅       ̅  are the sample means , S1
2
 and S2
2
 are the sample variances,n1 and n2 are the 
sample sizes  of group 1 and group 2 respectively.
1)  
A)  Suppose Xi denotes jobs created in the ith country. 
Then the sample average is, 
 ̅= 
 
 
  Xi = 425, 
The Standard deviation is, 
S = √
 
 
       ̅    =32.43455
B)  Here the hypothesis of interest is, 
H0: µ≤450 against H1: µ>450 
Where µ is the population mean number of jobs created in 2010. 
C)  To test this hypothesis we use the test statistic, 
t= 
 ̅    
    ̅ 
 which follows a t distribution with (n-1)=(5-1)=4 degrees of freedom 
under the null hypothesis. (SE( ̅) =Standard deviation/√ ) 
The value of the test statistic is, 
t= (425-450)/( 32.43455/√ ) = -1.7235 
D)   The p-value can be computed as follows, 
p-value = P(t > -1.7235); t follows  a t4 distribution. 
    =0.92 
Now at 95% confidence the value of significance level α=0.05. We reject the null 
hypothesis when the p-value is smaller than α. Now here we can see at 95% 
confidence the p-value, 0.92, is greater than α=0.05. Thus we are failing to reject 
the null hypothesis and we are concluding that the data gives enough evidence to 
conclude “There has been a significant decrease in the average no of jobs 
created.”
2)  
A) The standard error of mean is  
SE( ̅)=Standard deviation/√            =4/√    =0.25
B) Here the hypothesis of interest is, 
       H0: µ=30 against H1: µ>30;where µ is the population average of paper thickness. 
Here the test statistic, 
t= 
 ̅   
    ̅ 
 follows a t distribution with 255 degrees of freedom under the null 
hypothesis.
Since it is a one tail test so using the critical value approach we reject the null 
hypothesis at 100α% significance level or 100(1-α) % confidence if, 
Observed t> t255,α. 
Now, 
Observed t= 
       
    
 = 1.2 
Now at 95% confidence the tabulated critical value is, 
 t255,0.025= 1.65.
Since the observed value of the test statistic is smaller than the critical value so we are 
failing to reject the null hypothesis and concluding that at 95% confidence the data 
provides sufficient evidence to claim the mean thickness of the paper is not is 
significantly more than 30 mils.
C) Now the p-value is, 
p-value =P(t>1.2)= 0.11562651 
So as we can see the p-value is also greater than α=0.05 so we are failing to reject the 
null hypothesis.

it is an 8 question test that I have uploaded from a scanned document. if there are any questions please call

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