Question

TII a) through (1) below According to work done by a university professor, the proportion of individuals who cover their mouth wh Click here to view the standard normal distribution table (page 1). Click here to view the (a) What are the null and alternative hypotheses? Standard Normal Distribution Table (page 1) - X Standard Normal Distribution Table (page 2) HA Tube integers or decimals. Do not round. (b) Verify the requirements that allow use of the normal model to np. (1-p.)=Round to one lecimal place as needed.) A 0.00 1.116 1. 29 (c) Determine the sample proportion, 6=(Round to two decimal places as needed, (d) Determine the test statistic, 2 za (Round to two decimal places as needed.) Tel Determine the P-value P-value = (Round to three decimal places as needed. (f) State your conclusion There sufficient evidence at any commonly u 10.02 un Print Done Print Done

Answer 1

Here we have n=100 X=78 X 78 Ô == =0.78 n 100 a) Hypothesis Ho: P = 0.733 Ha: P = 0.733 This is a two tailed test. b)nP.(1 -P.) = 17.2 The sample size is less than 5% of the population size. X c) 78 - 100 =0.78 n d)Test statistics 0.78-0.733 P-P P(1-P)/n 0.733(1-0.733 100

= 1.06 e) P-value Pvalue at Z= 1.06 = 2 *(P(Z > 1.06)) w = 2*(1 - P(Z < 1.06)) = 2(1 – 0.855428) 0.289144 = 0.289 (Round to three decimal places) Decision Since P-value is greater than a (significance level) so, fail reject the null hypothesis. Conclusion There is not sufficient evidence at any commonly used level of significance a to conclude the proportion of individuals who cover their mouth when sneezing is not different from 0.733.