Question

part a) x = 137, s = 14.2, n = 20, H0: μ = 132, Ha: μ ≠ 132, α = 0.1

A) Test statistic: t = 1.57. Critical values: t = ±1.645. Do not reject H0. There is not sufficient evidence to conclude that the mean is different from 132.

B) Test statistic: t = 1.57. Critical values: t = ±1.729. Do not reject H0. There is not sufficient evidence to conclude that the mean is different from 132.

C) Test statistic: t = 0.35. Critical values: t = ±1.645. Do not reject H0. There is not sufficient evidence to conclude that the mean is different from 132.

D) Test statistic: t = 0.35. Critical values: t = ±1.729. Do not reject H0. There is not sufficient evidence to conclude that the mean is different from 132.

part b) A local group claims that the police issue more than 60 speeding tickets a day in their area. To prove their point, they randomly select two weeks. Their research yields the number of tickets issued for each day. The data are listed below. At α = 0.01, test the group's claim using P-values.

70 48 41 68 69 55 70 57 60 83 32 60 72 58

A) P-value = 0.4766. Since the P-value is great than α, there is not sufficient evidence to support the the group's claim.

B) P-value = 0.4766. Since the P-value is great than α, there is sufficient evidence to support the the group's claim.

part c) A local school district claims that the number of school days missed by its teachers due to illness is below the national average of μ = 5. A random sample of 28 teachers provided the data below. At α = 0.05, test the district's claim using P-values.

0 3 6 3 3 5 4 1 3 5 7 3 1 2 3 3 2 4 1 6 2 5 2 8 3 1 2 5

A) standardized test statistic ≈ -4.522; Therefore, at a degree of freedom of 27, P must lie between 0.0001 and 0.00003. P < α, reject H0. There is sufficient evidence to support the school district's claim.

B) standardized test statistic ≈ -4.522; Therefore, at a degree of freedom of 27, P must lie between 0.0001 and 0.00003. P < α, reject H0. There is no sufficient evidence to support the school district's claim

part d) To test the effectiveness of a new drug designed to relieve pain, 200 patients were randomly selected and divided into two equal groups. One group of 100 patients was given a pill containing the drug while the other group of 100 was given a placebo. What can we conclude about the effectiveness of the drug if 62 of those actually taking the drug felt a beneficial effect while 41 of the patients taking the placebo felt a beneficial effect? Use α = 0.05.

A) claim: p1 = p2; critical values z0 = ±1.96; standardized test statistic t ≈ 2.971; reject H0; The new drug is effective.

B) claim: p1 = p2; critical values z0 = ±1.96; standardized test statistic t ≈ 2.971; do not reject H0; The new drug is not effective.

Answer 1

a)

Ho :   µ =   132                  
Ha :   µ ╪   132       (Two tail test)          
                          
Level of Significance ,    α =    0.100                  
sample std dev ,    s =    14.2000                  
Sample Size ,   n =    20                  
Sample Mean,    x̅ =   137.0000                  
                          
degree of freedom=   DF=n-1=   19                  
                          
Standard Error , SE = s/√n =   14.2000   / √    20   =   3.1752      
t-test statistic= (x̅ - µ )/SE = (   137.000   -   132   ) /    3.1752   =   1.575
                          
critical t value, t* =    ±   1.729   

B) Test statistic: t = 1.57. Critical values: t = ±1.729. Do not reject H0. There is not sufficient evidence to conclude that the mean is different from 132.

======================

b)

Ho :   µ =   60                  
Ha :   µ >   60       (Right tail test)          
                          
Level of Significance ,    α =    0.010                  
sample std dev ,    s = √(Σ(X- x̅ )²/(n-1) ) =   13.4289                  
Sample Size ,   n =    14                  
Sample Mean,    x̅ = ΣX/n =    60.2143                  
                          
degree of freedom=   DF=n-1=   13                  
                          
Standard Error , SE = s/√n =   13.4289   / √    14   =   3.5890      
t-test statistic= (x̅ - µ )/SE = (   60.214   -   60   ) /    3.5890   =   0.060
                          
critical t value, t* =        2.650   [Excel formula =t.inv(α/no. of tails,df) ]              
                          
p-Value   =   0.4766   [Excel formula =t.dist(t-stat,df) ]              
Decision:   p-value>α, Do not reject null hypothesis                       

A) P-value = 0.4766. Since the P-value is great than α, there is not sufficient evidence to support the the group's claim.

=====================