Question

Boys of a certain age in the nation have an average weight of 86 with a standard deviation of 10.5 lb. A complaint is made that boys are overfed fed in a municipal children's home. As evidence, a sample of 19 boys of the given age is taken from the children's home with an average weight of 79.52 lb. What can be concluded with α = 0.05?

a) What is the appropriate test statistic?
---Select--- na z-test one-sample t-test independent-samples t-test related-samples t-test

b)
Population:
---Select--- boys in the nation weight children's home feeding method boys from the children's home
Sample:
---Select--- boys in the nation weight children's home feeding method boys from the children's home

c) Obtain/compute the appropriate values to make a decision about H₀.
(Hint: Make sure to write down the null and alternative hypotheses to help solve the problem.)
critical value =  ; test statistic =
Decision:  ---Select--- Reject H0 Fail to reject H0

d) If appropriate, compute the CI. If not appropriate, input "na" for both spaces below.
[  ,  ]

e) Compute the corresponding effect size(s) and indicate magnitude(s).
If not appropriate, input and select "na" below.
d =  ;   ---Select--- na trivial effect small effect medium effect large effect
r² =  ;   ---Select--- na trivial effect small effect medium effect large effect

f) Make an interpretation based on the results.

The weight of boys in the children's home was significantly higher than the weight of boys in the nation.The weight of boys in the children's home was significantly lower than the weight of boys in the nation.    The weight of boys in the children's home was not significantly different than the weight of boys in the nation.

Answer 1

A) Z TEST

BECAUSE σ is known

b)

population: boys in the nation

Sample: boys from the children's home

c)

Ho :   µ ≥ 86                  
Ha :   µ <   86       (Left tail test)          
                          
Level of Significance ,    α =    0.050                  
population std dev ,    σ =    10.5000                  
Sample Size ,   n =    19   110.2500              
Sample Mean,    x̅ =   79.5200                  
                          
'   '   '                  
                          
Standard Error , SE = σ/√n =   10.5000   / √    19   =   2.4089      
Z-test statistic= (x̅ - µ )/SE = (   79.520   -   86   ) /    2.4089   =   -2.690
                          
critical z value, z* =       -1.645   [Excel formula =NORMSINV(α/no. of tails) ]              
                          
Decision: z stat < -1.645, Reject null hypothesis                       

d)

Level of Significance ,    α =    0.05          
'   '   '          
z value=   z α/2=   1.960   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   10.5000   / √   19   =   2.4089
margin of error, E=Z*SE =   1.9600   *   2.4089   =   4.7213
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    79.52   -   4.721289   =   74.7987
Interval Upper Limit = x̅ + E =    79.52   -   4.721289   =   84.2413
95%   confidence interval is (   74.80   < µ <   84.24   )

e)

Cohen's d=|(mean - µ )/std dev|=   0.617 (large)
  
r² = d²/(d² + 4) =    0.087 (small)

f)

The weight of boys in the children's home was significantly lower than the weight of boys in the nation