Question

PART 1:

When bonding teeth, orthodontists must maintain a dry field. A new bonding adhesive has been developed to eliminate the necessity of a dry field. However, there is a concern that the new bonding adhesive is not as strong as the current standard, a composite adhesive. Tests on a sample of 26 extracted teeth bonded with the new adhesive resulted in a mean breaking strength (after 24 hours) of 2.33 MPa, and a standard deviation of 4 MPa. Orthodontists want to know if the true mean breaking strength is less than 4.06 MPa, the mean breaking strength of the composite adhesive. Assume normal distribution for breaking strength of the new adhesive.

1. The p-value of the test is (answer to 4 decimal places):

PART 2:

Are medical students more motivated than law students? A randomly selected group of each were administered a survey of attitudes toward Life, which measures motivation for upward mobility. The scores are summarized below. The researchers suggest that there are occupational differences in mean testosterone level. Medical doctors and university professors are two of the occupational groups for which means and standard deviations are recorded and listed in the following table.

Group	Sample size	Mean	StDev
Medical	n1 = 7	[(x)]1 = 81.59	s1 = 4.36
Law	n2 = 7	[(x)]2 = 76.27	s2 = 14.84

Let us denote:

• μ₁: population mean testosterone among medical doctors,
• μ₂: population mean testosterone among university professors,
• σ₁: population standard deviation of testosterone among medical doctors,
• σ₂: population standard deviation of testosterone among university professors.

1. If the researcher is interested to know whether the mean testosterone level among medical doctors is higher than that among university professors, what are the appropriate hypotheses he should test?
H₀: μ₁ = μ₂   against   H_a: μ₁ < μ₂.
H₀: μ₁ = μ₂   against   H_a: μ₁ > μ₂.
H₀: [(x)]₁ = [(x)]₂   against   H_a: [(x)]₁ ≠ [(x)]₂.
H₀: [(x)]₁ = [(x)]₂   against   H_a: [(x)]₁ > [(x)]₂.
H₀: [(x)]₁ = [(x)]₂   against   H_a: [(x)]₁ < [(x)]₂.
H₀: μ₁ = μ₂   against   H_a: μ₁ ≠ μ₂.

Case 1: Assume that the population standard deviations are unequal, i.e. σ₁ ≠ σ₂.
1. What is the standard error of the difference in sample mean [(x)]₁ − [(x)]₂? i.e. s.e.([(x)]₁−[(x)]₂) = [answer to 4 decimal places]

2. Rejection region: We reject H₀ at 10% level of significance if:
t < −1.89.
t > 1.41.
t < −1.41.
|t| > 1.89.
t > 1.89.
None of the above.

3. The value of the test-statistic is: Answer to 3 decimal places.

Case 2: Now assume that the population standard deviations are equal, i.e. σ₁ = σ₂.
1. Compute the pooled standard deviation, s_pooled [answer to 4 decimal places]

2. Rejection region: We reject H₀ at 10% level of significance if:
t > 1.78.
t < −1.36.
t > 1.36.
|t| > 1.78.
t < −1.78.
None of the above.

3. The value of the test-statistic is: Answer to 3 decimal places.

Answer 1

part 1)

Ho :µ =4.06     

Ha :µ <4.06(Left tail test)

  

Level of Significance , α = 0.05

sample std dev ,  s = 4.0000

Sample Size ,n = 26

Sample Mean,  x̅ =2.3300

  

degree of freedom=DF=n-1=25

  

Standard Error , SE =    s/√n =4.0000/ √ 26=0.7845

t-test statistic= (x̅ - µ )/SE = (2.330-4.06) / 0.784=-2.2053

  

p-Value=0.0184 [Excel formula =t.dist(t-stat,df) ]

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

part 2)

1.)

H0: μ1 = μ2   against   Ha: μ1 > μ2

CASE I

1)

Sample #1---->sample1

mean of sample 1,               x̅1=81.59

standard deviation of sample 1,s1 = 4.36

size of sample 1,                     n1=7

  

Sample #2---->sample2

mean of sample 2,               x̅2=76.270

standard deviation of sample 2,s2 = 14.84

size of sample 2,                     n2=7

  

difference in sample means = x̅1-x̅2   = 81.590-76.2700=5.3200

  

std error , SE = √(s1²/n1+s2²/n2) = 5.8461

----------------------------

2)

t-critical value , t* = 1.4149 (excel function:     =t.inv(α,df)

Rejection region: We reject H₀ at 10% level of significance if:
t > 1.41.

--------------------------------

3)

std error , SE = √(s1²/n1+s2²/n2) = 5.8461

t-statistic   = ((x̅1-x̅2)-µd)/SE   = (5.3200/5.8461) =0.910

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

ASSUMING equal variance

1)

pooled std dev ,    Sp=√([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) = 10.94

2)

Degree of freedom,    DF=n1+n2-2 = 12

t-critical value , t* = 1.36    (excel function:     =t.inv(α,df)

Rejection region: We reject H₀ at 10% level of significance if:

t > 1.36.

3)

std error , SE = Sp*√(1/n1+1/n2) = 5.8461

t-statistic   = ((x̅1-x̅2)-µd)/SE   = (5.3200-0) / 5.85=0.910