Question

A random sample of 80 patients was selected for an experimental treatment of temporomandibular aw) dysfunction (TMD) Prior to treatment. the patients filled out a survey on two nonfunctional jaw habits-bruxism (teeth grinding) and teeth clenching--that have been linked to TMD. Of the 80 patients, 8 admitted to only bruxism, 11 admitted to only teeth clenching, 32 admitted to both habits, and 29 claimed they had neither habit. Complete parts a through h below. e. Calculate the appropriate test statistic. 2 f. Give the rejection region for the test at (Round to three decimal places as needed.) 0.05. Reject Ho 2 Round to three decimal places as needed.) g. Give the appropriate conclusion in the words of the problem. Choose the correct conclusion below A. Do not reject Ho There is insufficient evidence that the percentages B Do not reject Ho There is sufficient evidence that the percentages associated with the admitted habits are not the same. associated with the admitted habits are not the same ) D. Reject Ho There is insufficient evidence that the percentages Reject H0. There is sufficient evidence that the percentages associated with the admitted habits are not the same C. associated with the admitted habits are not the same h. Find and interpret a 95% confidence interval for the true proportion of dental patients who admit to both habits Round to two decimal places as needed.) Interpret the results For a randomly selected patient, they will admit to both habits a percentage of the time between the confidence interval limits B. There is 95% confidence that the true population proportion of dental patients who admit to both habits is between the confidence interval limits. c. For a randomly selected sample of 60 patients, 95% of the time the proportion of patients who admit to bot habits will be between the confidence interval limits

Answer 1

Solution:-

observed Expected Num Category Num 20 7.2 Teeth clenching Both habits Neither habits Sum 20 4.05 32 20 7.2 29 20 4.05 80 80

e)

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

Null hypothesis: p₁= p₂ = p₃ = p₄

Alternative hypothesis: At least one of the proportions in the null hypothesis is false.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a chi-square goodness of fit test of the null hypothesis.

Analyze sample data. Applying the chi-square goodness of fit test to sample data, we compute the degrees of freedom, the expected frequency counts, and the chi-square test statistic. Based on the chi-square statistic and the degrees of freedom, we determine the P-value.

DF = k - 1 = 4 - 1
D.F = 3
(E_i) = n * p_i

$\chi ^{2}=\sum \left [\frac{(O_{r,c}-E_{r,c})^{2}}{E_{r,c}} \right ]$
X² = 22.5

where DF is the degrees of freedom, k is the number of levels of the categorical variable, n is the number of observations in the sample, E_i is the expected frequency count for level i, O_i is the observed frequency count for level i, and X² is the chi-square test statistic.

f) X²_Critical = 7.815

Reject H0, if X² > 7.815

Interpret results. Since the X²  value lies in the rejection region, hence we cannot accept the null hypothesis.

h) 95% confidence interval for the true proportion of dental patients who admits to both habits is C.I = ( 0.29, 0.51).
$C.I = \hat{p}\pm z_{\alpha /2}\times \sqrt{\frac{p(1-p)}{n}}$

C.I = 0.40 + 1.96 × 0.05477

C.I = 0.40 + 0.1074

C.I = ( 0.2926, 0.5074)