When bonding teeth, orthodontists must maintain a dry field. A new bonding adhes

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 10 extracted teeth bonded with the new adhesive resulted in a mean breaking strength (after 24 hours) of 2.65 MPa, and a standard deviation of 3.7 MPa. Orthodontists want to know if the true mean breaking strength is less than 5.22 MPa, the mean breaking strength of the composite adhesive. Assume normal distribution for breaking strength of the new adhesive. What are the appropriate hypotheses one should test? Ho: H5.22 against Ha: H 5.22 Ho: H2.65 against Ha H> 2.65 H0: ?-5.22 against Ha: ? > 5.22 0H0: ?-5.22 against Ha: ? 2.685 Submit Answer Tries 0/3 The value of the test-statistic is (answer to 3 decimal places): Tries 0/5 If a 0.025, what will be your conclusion? Submit Answer Do not reject Ho. Reject Ho- There is not information to conclude. Submit Answer Tries 0/2 The p-value of the test is (answer to 34decimal places): Submit Answer Tries 0/5 We should reject Ho for all significance level which are larger than p-value smaller than p-value.

Explanation / Answer

Solution:-

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

Null hypothesis: u > 5.22
Alternative hypothesis: u < 5.22

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.025. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), z statistic test statistic (z).

SE = s / sqrt(n)

S.E = 1.17004

z = (x - u) / SE

z = - 2.19

zcritical = - 1.96

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.

The observed sample mean produced a z statistic test statistic of -2.19.

Thus the P-value in this analysis is 0.014.

Interpret results. Since the P-value (0.014) is less than the significance level (0.025), we have to reject the null hypothesis.

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