Question

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 new bonding adhesive is not as strong as the current standard, a composite adhesive. ( Trends in Biomaterials & Artificial Organs, Jan. 2003.) Tests on a sample of 10 extracted teeth bonded with the new adhesive resulted in a mean breaking strength (after 24 hours) of x = 5.07 Mpa and a standard deviation of s = 0.46 Mpa, where Mpa = megapascal, a measure of force per unit area. Orthodontists want to know if the true mean breaking strength of the new bonding adhesive is less than 5.70 Mpa, the mean breaking strength of the composite adhesive. Conduct the appropriate analysis for the orthodontists. Use alpha= 0.01. What is the p-value?

Answer 1

`t=(5.07-5.7)/((0.46)/(√(10)))=-4.33`

`p_v =P(t_((9))<-4.33)=0.00095`

If we compare the p value and the significance level given
`\alpha=0.01` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, we can conclude that the true mean is less than 5.7 Mpa at 1% of signficance.

Explanation:

Data given and notation

`\bar X=5.07` represent the sample mean

`s=0.46` represent the sample standard deviation

`n=10` sample size

`\mu_o =5.7` represent the value that we want to test

`\alpha=0.01` represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

`p_v` represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is less than 5.7 Mpa, the system of hypothesis would be:

Null hypothesis:
`\mu \geq 5.7`

Alternative hypothesis:
`\mu < 5.7`

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

`t=(5.07-5.7)/((0.46)/(√(10)))=-4.33`

P-value

The first step is calculate the degrees of freedom, on this case:

`df=n-1=10-1=9`

Since is a one side lower tailed test the p value would be:

`p_v =P(t_((9))<-4.33)=0.00095`

Conclusion

If we compare the p value and the significance level given
`\alpha=0.01` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, we can conclude that the true mean is less than 5.7 Mpa at 1% of signficance.