Question

An attorney claims that more than 25% of lawyers advertise their services. A sample of 200 lawyers in a certain city showed that 63 had used some form of advertising a. At a 0.05, is there enough evidence to support the attorneys claim? Sample information 2 points Hypothesis information Ho: на: Test statistics p-value Conclusion: 3 points 3 points 3 points 3 points b. Find a 95% confidence interval for the true proportion of attorneys who advertise. Error Bound Calculation Confidence interval Conclusion: 3 points 3 points 3 points 3 points In a nursing home study, the researchers found that 12 out of 34 small nursing homes had a resident vaccination rate of less than 80%, while 17 out of 24 large nursing homes had a resident vaccination rate of less than 80%. a. At a=0.05, test the claim that there is no difference in the proportions of small and large nursing homes with a resident vaccination rate of less than 80%. Sample information 2 points Hypothesis information Ho: Ha: Test statistics p-value Conclusion: 3 points 3 points 3 points 3 points

Answer 1

In line with Chegg's guidelines answering the first question 3

a)

One-Proportion Z test

The following information is provided: The sample size is N = 200, the number of favorable cases is X = 63 and the sample proportion is pˉ​=X/N​=63/200​=0.315, and the significance level is α=0.05 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: p =0.25 Ha: p >0.25 This corresponds to a Right-tailed test, for which a z-test for one population proportion needs to be used. (2a) Critical Value Based on the information provided, the significance level is α=0.05, therefore the critical value for this Right-tailed test is Zc​=1.6449. This can be found by either using excel or the Z distribution table. (2b) Rejection Region The rejection region for this Right-tailed test is Z>1.6449 (3) Test Statistics The z-statistic is computed as follows: $Z = \frac{\bar p - p_0}{\sqrt{p_0(1-p_0)/n}} = \frac{ 0.315 - 0.25 }{\sqrt{ 0.25(1- 0.25)/100}} = 2.1229$ (4) The p-value The p-value is the probability of obtaining sample results as extreme or more extreme than the sample results obtained, under the assumption that the null hypothesis is true. In this case, the p-value is p =P(Z>2.1229)=0.0169 (5) The Decision about the null hypothesis (a) Using the traditional method Since it is observed that Z=2.1229 > Zc​=1.6449, it is then concluded that the null hypothesis is rejected. (b) Using p-value method Using the P-value approach: The p-value is p=0.0169, and since p=0.0169≤0.05, it is concluded that the null hypothesis is rejected. (6) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population proportion p is greater than 0.25, at the 0.05 significance level.

b)

One-Proportion Confidence Interval

We need to construct the 95% confidence interval for the population proportion. We have been provided with the following information: The sample size is N = 200, the number of favorable cases is X = 63 and the sample proportion is pˉ​=X/N​=63/200​=0.315, and the significance level is α=0.05 Critical Value Based on the information provided, the significance level is α=0.05, therefore the critical value for this Two-tailed test is Zc​=1.96. This can be found by either using excel or the Z distribution table. Margin of Error $M.E =1.96 \times \sqrt{\frac{0.315(1- 0.315)}{200}}=0.0644$ The confidence interval: $\begin{array}{ccl} CI(\text{Proportion}) \displaystyle \left( \hat p - z_c \sqrt{\frac{\hat p (1-\hat p)}{n}}, \hat p + z_c \sqrt{\frac{\hat p (1-\hat p)}{n}} \right) \\ \\ = \displaystyle \left( 0.315 - 1.96 \times \sqrt{\frac{0.315(1- 0.315)}{200}}, 0.315 + 1.96 \times \sqrt{\frac{0.315(1- 0.315)}{200}} \right) \\ \\ = (0.2506, 0.3794) \end{array}$ Therefore, based on the data provided, the 95% confidence interval for the population proportion is 0.2506<p<0.3794, which indicates that we are 95% confident that the true population proportion p is contained by the interval (0.2506,0.3794)

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