Question

A new dental bondlng agent. When bonding teeth, orthodontists must maintain a dry field. A new bonding adhesive(called "Smartbond") has been developed to eliminate the necessity of a dry field. However, there is concern that the newbonding adhesive is not as strong as the current standard, a composite adhesive. (Trends in Biomaten'a/s & ArtificialOrgans, January 2003.) Tests on a sample of 10 extracted teeth bonded with the new adhesive resulted in a meanbreaking strength (after 24 hours) of} = 5,07 Mpa and a standard deviation of s = .46 Mpa, where Mpa = megapascal, ameasure of force per unit area. Orthodontists want to know if the true mean breaking strength of the new bondingadhesive is less than 5.70 Mpa, the mean breaking strength of the composite adhesive.3. Set up the null and alternative hypotheses for the test.b. Find the rejection region for the test, using a = .01.c. Compute the test statistic.d. State the appropriate conclusion for the teste. What conditions are required for the test results to be valid?

Answer 1

(a) H₀: μ ≥ 5.70 vs. Hₐ:μ < 5.70

(b) The rejection region is (t₀.₀₁,₉ ≤ -2.821).

(c) The value of the test statistic is -4.33.

(d) The true mean breaking strength of the new bonding adhesive is less than 5.70 Mpa.

Explanation:

A hypothesis test should be conducted to determine that the if the true mean breaking strength of the new bonding adhesive is less than 5.70 Mpa.

(a)

The hypothesis is:

H₀: The true mean breaking strength of the new bonding adhesive is not less than 5.70 Mpa, i.e. μ ≥ 5.70.

Hₐ: The true mean breaking strength of the new bonding adhesive is less than 5.70 Mpa, i.e. μ < 5.70.

(b)

The alternate hypothesis indicates that the hypothesis test is left-tailed.

The rejection region for the left tailed test will be towards the lower tail of the t-distribution curve.

The significance level of the test is: α = 0.01.

The critical value is:

`t_(\alpha ,(n-1))=t_(0.01,(10-1))=t_(0.01,9)`

Use the t-table for the critical value.

`t_(\alpha ,(n-1))=t_(0.01,9)=-2.821`

Since rejection region is in the lower tail the critical value will be negative.

Thus, the rejection region is (t₀.₀₁,₉ ≤ -2.821).

(c)

The test statistic value is:

`t=(\bar x-\mu)/(s/√(n))`

Given:

`\bar x=5.07\\s=0.46\\n=10\\\mu=5.70`

Compute the value of the t-statistic as follows:

`t=(\bar x-\mu)/(s/√(n))=(5.07-5.70)/(0.46/√(10)) =-4.33`

The value of the test statistic is -4.33.

(d)

The value of the test is less than the critical value.

`t=-4.33<t_(0.01,9)=-2.821`

This implies that the test statistic lies in the rejection region.

Hence the null hypothesis will be rejected at 1% significance level.

Conclusion:

As the null hypothesis is rejected it can be concluded that the true mean breaking strength of the new bonding adhesive is less than 5.70 Mpa.

(e)

The conditions required for the t-test for single mean to be valid is:

• The data should be continuous.
• The parent population should be normally distributed.
• The sample should be randomly selected.