Question

Let μ₁ and μ₂ denote true average densities for two different types of brick. Assuming normality of the two density distributions, test:

H₀: μ₁ − μ₂ = 0 versus H_a: μ₁ − μ₂ ≠ 0

Use the following data:
n₁ = 6, x₁ = 24.71, s₁ = 0.53, n₂ = 5, x₂ = 21.85, and s₂ = 0.530.

The conservative degrees of freedom, min(n1-1, n2-1), are:

df =

Use α = 0.05. Round your test statistic to three decimal places and your P-value to four decimal places.)
Use the degrees of freedom above for your p-value.
Recall the command pt(t, df) will provide the area under the t distribution. Manipulate this to find the appropriate area under the curve that represents the p-value from your test.

t =
P-value =

State whether or not to reject the null.

Reject H₀.Fail to reject H₀.     

State the conclusion in the problem context.

The data provides moderately suggestive evidence of a difference between the true average densities for the two different types of brick.The data provides no evidence of a difference between the true average densities for the two different types of brick.     The data provides convincing evidence of a difference between the true average densities for the two different types of brick.The data provides suggestive, but inconclusive evidence of a difference between the true average densities for the two different types of brick.

The point estimate and margin of error for the 95% confidence interval for μ₁ − μ₂ are: (Use the minimum degrees of freedom for your critical value.)
±
You may need to use the t-table to complete this problem.

Answer 1

DF = min(n1-1 , n2-1 )=   4
----------------

Ho :   µ1 - µ2 =   0          
Ha :   µ1-µ2 ╪   0          
                  
Level of Significance ,    α =    0.05          
                  
Sample #1   ---->   1          
mean of sample 1,    x̅1=   24.71          
standard deviation of sample 1,   s1 =    0.53          
size of sample 1,    n1=   6          
                  
Sample #2   ---->   2          
mean of sample 2,    x̅2=   21.850          
standard deviation of sample 2,   s2 =    0.53          
size of sample 2,    n2=   5          
                  
difference in sample means = x̅1-x̅2 =    24.710   -   21.8500   =   2.8600
                  
std error , SE =    √(s1²/n1+s2²/n2) =    0.3209          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   2.8600   /   0.3209   ) =   8.912

p-value =    0.0009

Decision:   p value < α , Reject Ho,   

The data provides convincing evidence of a difference between the true average densities for the two different types of brick.

DF = min(n1-1 , n2-1 )=   4              
t-critical value , t* =    2.7764   (excel formula =t.inv(α/2,df)          
                  
                  
                  
std error , SE =    √(s1²/n1+s2²/n2) =    0.321          
margin of error, E = t*SE =    2.7764   *   0.321   =   0.8910
                  
point estimate = difference of means = x̅1-x̅2 =    24.7100   -   21.850   =   2.860
confidence interval is                   
Interval Lower Limit = point estimate  ± E = 2.860 ± 0.8910