When bonding teeth, orthodontists must maintain a dry field. A new bonding adhes
ID: 3243830 • Letter: W
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 2.77 MPa, and a standard deviation of 4.1 MPa. Orthodontists want to know if the true mean breaking strength is less than 4.38 MPa, the mean breaking strength of the composite adhesive. Assume normal distribution for breaking strength of the new adhesive.
The value of the test-statistic is -2.002
1. The p-value of the test is?
Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: > 4.38
Alternative hypothesis: < 4.38
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 z-test.
Analyze sample data. Using sample data, we compute the standard error (SE), and the z statistic test statistic (z).
SE = s / sqrt(n)
S.E = 0.8041
z = (x - ) / SE
z = - 2.0002
where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.
The observed sample mean produced a z statistic test statistic of - 2.002. We use the z Distribution Calculator to find P(z > - 2.002) = 0.0228.
Thus the P-value in this analysis is 0.0228.
Interpret results. Since the P-value (0.0228) is greater than the significance level (0.01), we cannot reject the null hypothesis.
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