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 27 extracted teeth bonded with the new adhesive resulted in a mean breaking strength (after 24 hours) of 2.61 MPa, and a standard deviation of 3.9 MPa. Orthodontists want to know if the true mean breaking strength is less than 4.64 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=4.64H0:μ=4.64 against Ha:μ>4.64Ha:μ>4.64 .
H0:μ=2.61H0:μ=2.61 against Ha:μ≠2.61Ha:μ≠2.61 .
H0:μ=2.61H0:μ=2.61 against Ha:μ>2.61Ha:μ>2.61 .
H0:μ=4.64H0:μ=4.64 against Ha:μ≠4.64Ha:μ≠4.64 .
H0:μ=4.64H0:μ=4.64 against Ha:μ<4.64Ha:μ<4.64 .
H0:μ=2.61H0:μ=2.61 against Ha:μ<2.61Ha:μ<2.61 .

2. The formula of the test-statistic to use here is
x¯−μ0s/n√x¯−μ0s/n .
x¯−μ0σ/n√x¯−μ0σ/n .
p^−p0p0(1−p0)/n−−−−−−−−−−√p^−p0p0(1−p0)/n .
None of the above.

3. Rejection region: We should reject H0H0 at 5% level of significance if:
test statistic >1.706>1.706 .
|test statistic| >1.960>1.960 .
test statistic >1.645>1.645 .
|test statistic| >2.056>2.056 .
test statistic <−1.706<−1.706 .
test statistic <−1.645<−1.645 .

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

...

5. If α=0.05α=0.05 , what will be your conclusion?
There is not information to conclude.
Do not reject H0H0 .
Reject H0H0 .

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

7. We should reject H0H0 for all significance level which are
larger than p-value.
smaller than p-value.
not equal to p-value.

Answer 1

here we have

and n = 27

1 ) H0:μ=4.64 against Ha:μ<4.64

2 ) since the population standard deviation is unknown so the formula used here

3 ) The rejection region for this left-tailed test is R=t:t < −1.706

4 ) t = = -2.705

t = -2.705

5 ) Since it is observed that t=−2.705 < t_c​=−1.706, it is then concluded that the null hypothesis is rejected. Reject H0

6 ) The p-value is p = .0059

7 ) since p=0.0059<0.05, it is concluded that the null hypothesis is rejected.

We should reject H0 for all significance level which are smaller than p-value.