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

1. What are the appropriate hypotheses one should test?
H₀: μ = 4.06 against  H_a: μ > 4.06.
H₀: μ = 4.06 against  H_a: μ ≠ 4.06.
H₀: μ = 2.33 against  H_a: μ ≠ 2.33.
H₀: μ = 2.33 against  H_a: μ > 2.33.
H₀: μ = 2.33 against  H_a: μ < 2.33.
H₀: μ = 4.06 against  H_a: μ < 4.06.

2. The formula of the test-statistic to use here is
\dfrac[(x)] − μ₀σ/√n.
\dfrac[(x)] − μ₀s/√n.
\dfrac[^(p)] −p₀√{p₀(1−p₀)/n}.
None of the above.

3. Rejection region: We should reject H₀ at2.5% level of significance if:
test statistic < −1.960.
|test statistic| > 2.241.
test statistic < −2.060.
test statistic > 1.960.
|test statistic| > 2.385.
test statistic > 2.060.

4. The value of the test-statistic is (answer to 3 decimalplaces):

5. If α = 0.025, what will be your conclusion?
Do not reject H₀.
Reject H₀.
There is not information to conclude.

6. The p-value of the test is (answer to 4 decimalplaces):

7. We should reject H₀ for all significancelevel which are
not equal to p-value.
larger than p-value.
smaller than p-value.

Are medical students more motivated than law students?A randomly selected group of each were administered a survey ofattitudes toward Life, which measures motivation for upwardmobility. The scores are summarized below. The researchers suggestthat there are occupational differences in mean testosterone level.Medical doctors and university professors are two of theoccupational groups for which means and standard deviations arerecorded 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 medicaldoctors,
• μ₂: population mean testosterone among universityprofessors,
• σ₁: population standard deviation of testosteroneamong medical doctors,
• σ₂: population standard deviation of testosteroneamong university professors.

1. If the researcher is interested to know whether the meantestosterone level among medical doctors is higher than that amonguniversity professors, what are the appropriate hypotheses heshould 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 deviationsare 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 decimalplaces.

4. If α = 0.1, and the p-value is 0.1965, what will be yourconclusion?
Do not reject H₀.
Reject H₀.
There is not enough information to conclude.

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

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 decimalplaces.

4. If α = 0.1, , and the p-value is 0.1904, what will be yourconclusion?
Reject H₀.
There is not enough information to conclude.
Do not reject H₀.