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RSMT 1501 Review for Final Exam

RSMT 1501 Review for Final Exam Page 1 1. A product manager in a company plans to bring designers from around the world for an ad campaign. To estimate the average time required to obtain work permits for designers, a sample of 36 applicants has taken which has a mean time of 118.2 days. Assume the population standard deviation is 18.1 days. (a) Find the point estimate for the average time to obtain work permit for all applicants. (b) Construct a 90% confidence interval for the population average time to obtain work permit. (c) Construct a 95% confidence interval for the population average time to obtain work permit. (d) Construct a 99% confidence interval for the population average time to obtain work permit. (e) Which interval (90%, 95%, or 99%) is the widest? (f) What is the sample size needed if the precision is to be within Â± 3.6 days with a 95% confidence level? 2. A marketing manager wants to know how potential car buyers rate the effectiveness of a new car design in persuading them to buy the car. Twelve of the buyers were asked to score the design in a scale of 1 to 10. The mean score of the result is 7 with a standard deviation of 2. Assume that score values of all potential car buyers are normally distributed: (a) Construct the 99% confidence interval for the mean score of all potential car buyers. (b) Construct the 90% confidence interval for the mean score of all potential car buyers and interpret it. 3. An e-commerce company claims there is an average of 2300 hits per minute in a website for promoting a new product. A random sample of 64 minutes in a business day has an average of 2240 hits. (a) Assuming that the population standard deviation is known as 120 hits, is there sufficient evidence to infer that the average number of hits on the website is different from 2300 per minute? Test at 1% level of significance. (b) Which type of error (I or II) might have occurred in your conclusion? Explain the error in that situation. 4. A vacation tour company advertised that the average cost of land excursions in the Caribbean is less than $80. A random sample of 25 excursions has an average cost of $76.50. Assume the population is normally distributed and the population standard deviation is $18.20. (a) Is there enough evidence that the average cost is less than $80 at the 5% level of significance? (b) Which type of error (I or II) might have occurred in your conclusion? Explain the error in that situation.RSMT 1501 Review for Final Exam Page 2 5. Myers Interior Design store tells customers that a special order will take at most 42 days. During recent months the owner has received several complaints that the special orders are taking longer than 42 days. A sample of 14 special orders delivered in the last month showed that the mean waiting time was 51 days, with a standard deviation of 8 days. At the 0.01 significance level, can you conclude that the mean waiting time for the special orders is longer than 42 days? Assume the population is normally distributed. 6. A statistics teacher wanted to see whether there was a significant difference in ages between day students and night students. A random sample of 35 students is selected from each group. The data are given below. Test the claim that there is a difference in mean ages between the two groups. Use ? = 0.05. From past studies, the population standard deviation for day studentsâ€™ group is known to be 3.1 years old and the population standard deviation for evening studentsâ€™ group is known to be 3.3 years old. Day Students 22 24 24 23 19 19 23 22 18 21 21 18 18 25 29 24 23 22 22 21 20 20 20 27 17 19 18 21 20 23 26 30 25 21 25 Use the information below to answer each question. z-Test: Two Sample for Means Day students Evening students Mean 22 23.29 Observations 35 35 Hypothesized Mean Difference 0 (a) State the null and alternative hypotheses. (Identify ?1 and ?2 before statements). (b) Find the critical value(s) and identify the rejection (of Ho) region(s). (c) What is the test statistic value? (d) Make a decision to reject or fail to reject the null hypothesis and interpret the decision. Page 3 7. One of the goals of promotion campaign is to improve customersâ€™ acceptance towards the brand. Eight customers were given a questionnaire about the brand before and after the promotion campaign. Higher score indicates higher rate of acceptance to the brand. Assume that acceptance scores of all customers are normally distributed. Use the information below and a 1% significance level to answer the following questions. t-Test: Paired Two Sample for Means After campaign Before campaign Observations 8 8 Hypothesized Mean Difference 0 df 7 (1) Using the output above, can you conclude that the campaign improves the acceptance? (a) State the null and alternative hypotheses. (b) Find the critical value(s) and identify the rejection (of Ho) region(s). (c) What is the test statistic value? (d) Make a decision to reject or fail to reject the null hypothesis and interpret the decision. (2) Using the output above, can you conclude that the campaign can change the acceptance? (a) State the null and alternative hypotheses. (b) Find the critical value(s) and identify the rejection (of Ho) region(s). (c) What is the test statistic value? (d) Make a decision to reject or fail to reject the null hypothesis and interpret the decision.