Carry your intermediate computations to at least three decim

Carry your intermediate computations to at least three decimal places and round your answers as specified in the table. (If necessary, consult a list of formulas.)

The null hypothesis:

H0:

The alternative hypothesis:

H1:

The type of test statistic:

(Choose one)

Z

t

Chi square

F

The value of the test statistic:
(Round to at least three decimal places.)

The p-value:
(Round to at least three decimal places.)

Can we reject the claim that the mean IQ score of new hires from the current year is greater than or equal to the mean IQ score of new hires from last year?

Yes

No

The human resources department of an engineering company gives IQ tests to a randomly selected group of new hires every year. They claimed that the mean IQ score of new hires, , from this year is greater than or equal to the mean IQ score of new hires, , from last year. This year, new hires took the test and scored an average of points with a standard deviation of . Last year, new hires took the IQ test and they scored an average of points with a standard deviation of . Assume that the population standard deviation of the IQ scores from the current year and the last year can be estimated by the sample standard deviations, since the samples that are used to compute them are quite large. Is there enough evidence to reject the claim of the human resources department, at the level of significance? Perform a one-tailed test. Then fill in the table below. Carry your intermediate computations to at least three decimal places and round your answers as specified in the table. (If necessary, consult a list of formulas.)

Solution

Formulating the null and alternative hypotheses,              
              
Ho:   u1 - u2   <=   0   [ANSWER]
Ha:   u1 - u2   >   0   [ANSWER]

*****************************

As n1 + n2 > 60, then we use Z STATISTIC. [ANSWER]

***************************

At level of significance =    0.05          

As we can see, this is a    right   tailed test.      
Calculating the means of each group,              
              
X1 =    113.1          
X2 =    116.9          
              
Calculating the standard deviations of each group,              
              
s1 =    16.2          
s2 =    17.7          
              
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):              
              
n1 = sample size of group 1 =    80          
n2 = sample size of group 2 =    75          
Also, sD =    2.730878979          
              
Thus, the z statistic will be              
              
z = [X1 - X2 - uD]/sD =    -1.391493372   [ANSWER, Z STATISTIC]

*********************************  
                           The p value for this one is
              
p =    0.082037934 [ANSWER, P VALUE]

******************************
      
              
Comparing this to the significance level,    WE FAIL TO REJECT THE NULL HYPOTHESIS.          

So, NO: WE CANNOT REJECT. [ANSWER, NO]