Question

An online clothing retailer knows that historically 20% of potential customers that visit its website actually buy something To see if they can increase the proportion of visitors who buy something, they randomly select 1000 visitors to the website and offer them free shipping if they make a purchase. Of the 1000 visitors who receive this offer, 230 actually buy something. The company wants to test to see if offering free shipping to all customers would increase the proportion who would buy something 2. a. Define the parameter and state the hypotheses. [3pts] b. Find the observed sample statistic and the test statistic. Use proper notation for the sample statistic (7 or p), and clearly identify your test statistic as either a z-score or a t-score. [4pts] C. Find the p-value of your test statistic. You are required to graph the test statistic on its corresponding distribution, shade the appropriate area, and provide correct calculator documentation. Box final answer. [3pts] d. Make a generic decision about H - [2pts] e. Make a conclusion in the context of the problem. [2pts] Explain your answer f. Do you have any concerns with this hypothesis test? Clearly answer"yes" or "no". and provide calculations. [2pt]

Answer 1

The following information is provided: The sample size is N 1000, the number of favorable cases is X = 230, and the sample proportion is p = x 0.23, and the significance level is a = 0.05 230 1000 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho:p=0.20 Ha: p > 0.20 This corresponds to a right-tailed test, for which a z-test for one population proportion needs to be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the critical value for a right-tailed test is ze = 1.64. The rejection region for this right-tailed test is R = {z:z > 1.64} (3) Test Statistics The Z-statistic is computed as follows: 2 = P - po po(1 – po)/n 0.23 – 0.20 V0.20(1 – 0.20)/1000 2.372 (4) Decision about the null hypothesis Since it is observed that z = 2.372 > 2c = 1.64, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p 0.0089, and since p = 0.0089 < 0.05, it is concluded that the null hypothesis is rejected. (5) 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 po, at the a = 0.05 significance level. Confidence Interval The 95% confidence interval for p is: 0.204 < p < 0.256. Graphically. Z-Test Results: z-stats = 2.372, p-value =0.0089 0.40 0.35 0.30 0.25 0.20 0.15 0.10 0.05 0.00 -4.0 -3.5 -3.0 -2.5 -2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

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