Question

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. 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 at 2.5% 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 decimal places):

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 decimal places):

7. We should reject H₀ for all significance level 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 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.

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

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.

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

Answer 1

Solution:-

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) H₀: μ = 4.06 against   H_a: μ < 4.06.

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u = 4.06
Alternative hypothesis: u < 4.06

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.01. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 0.7845
DF = n - 1

D.F = 25
2)

3)

t_Critical = - 2.06

Rejection region is test statistic < −2.060.

4)

t = - 2.21

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.

5)

Interpret results. Since the t-value (-2.21) lies in the rejection region, hence we have to reject the null hypothesis.

Reject H₀.

6) The p-value of the test is 0.018.

7) We should reject H₀ for all significance level which are larger than p-value.