Question

The residents of Flint, Michigan are hoping to show that lead in the pipes is lowering the IQ of children in the city and need to have a statistician give evidence at the p<.05 level to do so. They have a psychologist randomly administer IQ tests to 200 children in the town and they find the average IQ of those children is 98. The IQ of children in general is known to have a mean of 100 with a standard deviation of 15.

1. Restate the question:

   • Population 1:

   • Population 2:

   • Research hypothesis:

   • Null hypothesis:

   • Determine the characteristics of the comparison distribution to compare to sample distribution:

     • μ 1 (Population 1):

     • N (Population 1):

     • μ2 (Population 2):

     • σ2(Population 2):

     • μ2M (Population 2, distribution of means):

     • σ2M (Population 2, distribution of means):

     • Is the distribution of means normally distributed?

     • Why does the distribution of means have a normal distribution, a non-normal distribution, or an unknown distribution?

     • c. Determine the critical value (cutoff score) on the comparison distribution at which the null hypothesis should be rejected

     • Determine the Z score of your sample

     • Decide whether to reject the null hypothesis

     • State the findings in term of the study

     • Calculate the 95% confidence interval for what the true IQ is of children in Flint.

Answer 1

population 1: children of flint town

population 2 : overall children in general

research hypothesis : IQ of flint children is lower than overall children

null hypothesis : IQ of flint children is equal to overall chldren

• μ 1 (Population 1): 98

• N (Population 1): 200

• μ2 (Population 2): 100

• σ2(Population 2): 15

• the distribution of means have a normal distribution,

.................

critical z value, z* =       -1.6449   [Excel formula =NORMSINV(α/no. of tails) ]      
so, critical region :  

test stat < -1.6449

p-Value   =   0.0297
Decision:   p-value<α, Reject null hypothesis   

...

Ho :   µ =   100                  
Ha :   µ <   100       (Left tail test)          
                          
Level of Significance ,    α =    0.05                  
population std dev ,    σ =    15.0000                  
Sample Size ,   n =    200                  
Sample Mean,    x̅ =   98.0000                  
                                
Standard Error , SE = σ/√n =   15.0000   / √    200   =   1.0607      
Z-test statistic= (x̅ - µ )/SE = (   98.000   -   100   ) /    1.0607   =   -1.89

........

Z stat < critical value , reject Ho

..........

the IQ level of flint city children is less than children in general

............

Level of Significance ,    α =    0.05          
z value=   z α/2=   1.9600   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   15.0000   / √   200   =   1.060660
margin of error, E=Z*SE =   1.9600   *   1.06066   =   2.078856
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    98.00   -   2.078856   =   95.921144
Interval Upper Limit = x̅ + E =    98.00   -   2.078856   =   100.078856
95%   confidence interval is (   95.92   < µ <   100.08   )

thanks

please revert bakc for any doubt