Assignment 3 Final

Introduction to Quantitative Analysis Assignment #3 Parinya Showanasai:4949613 P318 9.46 Children in United States account directly for $36 billion in sales annually. When their indirect influence over product decisions from stereos to vacations is considered, the total economic spending impacted by children in the United States is $290 billion. It is estimated that by age 10, a child makes an average of over five trips a week to a store (M.E.Goldberg, G. J, Gorn, L.A. Peracchio, and G. Bamossy, “Understanding Material Among Youth,” Journal of Consumer Psychology, 2003, 13(3);278-288). Suppose that you want to prove that children in your city average more than five trips a week to a store. Let µ represent the population mean number of times children in your city make trips to store. a. State the null and alternative hypothesis. b. Explain in the context above scenario the meaning of type I and Type II errors c. Suppose that you carry out a study in the city in which you live. Based on past studies, you assume that the standard deviation of the number of trips to the store is 1.6. you take a sample of 100 children and find that the mean numbers of trips to the store is 5.47. At the 0.01 level of significance, is there evidence that the population mean number of trips to the store is greater than 5 per week? d. Interpret the meaning of the p-value in (c) a) The null hypothesis is that the population mean has not changed, the average number of children going to the store is 5 times a week. Ho: µ ≤ 5 The alternative hypothesis is that the population mean is not correct, using the samples in city, the population means should be more than 5 times a week Ha: µ > 5 b) In this study, the conclusion of the study from these samples can be wrong in two ways: 1- The null hypothesis is rejected when it is true (Type I error). This could happen when the samples that we surveyed might be in an extreme group (people who go shopping too often) that does not represent the whole population, this will make the sample mean to fall outside an acceptable area. The error also depends on the significant level specified. Probability of this error equals to significance level determined which is alpha (α). Im this case we reject null hypothesis when it is true

2- the samples in the survey do not represent population, and they are in another extreme group who go to the store less than 5 times a week. (beta β)

c) The study has a sample size of 100, sample standard deviation is 1.6 and the mean 5.47 is obtained. The question is, at 0.01 level of significance, if this sample represents the population or in other word, if children in this area go to the store more often than the average.  p-value = P[ p = 5.47 | p ≤ 5.0, n = 100] Using normal distribution and Z test statistics: Determine the rejection region: Z =( X -µ)/(σ/ √n) Z = (5.47-5)/(1.6/√100) = 0.47/0.16 = 2.9375

Assuming a normal distribution, the p-value can be calculated using excel To calculate using Z-score =NORMSINV(0.01) = -2.32634 =NORMSDIST(z-score) = 0.998345 Or 1- 0.998345 = 0.001654 = pvalue The p-value, 0.001654, Ho has to be rejected As the p-value is lower than the significant level, it means that observed level of significance is lower than actual level of significance. There is sufficient evidence that it is possible that the population mean might not be 5 or less, the statistics of the samples show 1) the p-value is less than 0.01 and 2) Z value of the test is greater than Z value of an upper bound(Z=2.9375 > Z =2.33)

P324: 9.59 In the New York State, saving banks are permitted to sell a form of life insurance called Savings Bank Life Insurance (SBLI). The approval process consists of underwriting , which includes a review of the application, a medical information bureau check, possible requests for additional medical information and medical exams, and a policy compilation stage where the policy pages are generated and sent to the bank for delivery. The ability to deliver approved policies to customers in a timely manner is a critical to the profitability of this service. During a period of one month, a random sample of 27 approved policies is selected INSURANCE and the total processing time in days recorded: a) In the past , the mean processing time averaged 45 days. At the 0.05 level of significance, is there evidence that the mean processing time has changed from 45 days? b) What assumption about the population distribution is needed in (a) c) Do you think that the assumption needed in (b) is seriously violated? Explain. d) ??? e) Suppose that you expect that the mean processing time has increased. Formulate a null and alternative hypothesis and give the p-value for this test, using SPSS output f) Suppose that you expect that the mean processing time has decreased. Formulate a null and alternative hypothesis and give the p-value for this test, using SPSS output

a) In the past , the mean processing time averaged 45 days. At the 0.05 level of significance, is there evidence that the mean processing time has changed from 45 days By running SPSS testing one sample statistics, using the median = 45, the test shows that the mean of the samples is not different from the value 45. (twotailed test significance level at 95%) and the null hypothesis: Ho: xbar ≠ 45 Cannot be rejected.

b) What assumption about the population distribution is needed in (a) The assumption of normality is usually used in analysis. c) Do you think that the assumption needed in (b) is seriously violated? Explain.

The assumption of normality is tested and shown that the distribution of samples is acceptable. d) ??? e) Suppose that you expect that the mean processing time has increased. Formulate a null and alternative hypothesis and give the p-value for this test, using SPSS output If we suspect that the mean of processing time has decreased, therefore the null and alternative hypothesis will be: Ho: X ≤ 45 Ha: X > 45 SPSS provides a one simple ttest of the samples comparing to the value 45. p-value can be calculated from the value labeled “Sig (2-tailed)”. This value is a area under the distribution curve started from the left. The sign of t-value is minus which means that the value is on the left of the mean. The p-value equals to 0.821 (starts from the right of the curve to the left). Therefore the null hypothesis cannot be rejected as p-value is much greater than confidence level (0.05) Also, SPSS shows that the upper level equals to 45+7.19 = 52.19 .The value 46.89 is in this range and the null hypothesis cannot be rejected

f) Suppose that you expect that the mean processing time has decreased. Formulate a null and alternative hypothesis and give the p-value for this test, using SPSS output If we suspect that the mean of processing time has decreased, therefore the null and alternative hypothesis will be: Ho: X ≥ 45 Ha: X < 45 The p-value equals to 0.179 (starts from the left to the right). Therefore the null hypothesis cannot be rejected as p-value is much greater than confidence level (0.05) Also, SPSS shows that the lower level equals to 45-9.41 = 33.59 .The value 43.89 is greater than a lower bound and the null hypothesis cannot be rejected.