Part B Course Project Math 533 Essay Example Essay Example

Part B Course Project Math 533 Essay Example Paper Part B Course Project Math 533 Essay Introduction Course Project Part B a. the average (mean) annual income was less than $50,000 Null and Alternative Hypothesis H0: mu= 50 (in thousands) Ha: mult;50 (in thousands) Level of Significance Level of Significance = . 05 Test Statistic, Critical Value, and Decision Rule Since alpha = . 05, zlt;-1. 645, which is lower tailed Rejection region is, zlt;-1. 645 Calculate test statistic, x-bar=43. 74 and s=14. 64 Z=(43. 74-50)/2. 070=-3. 0242. 070 is calculated by: s/sq-root of n Decision Rule: The calculated test statistic of -3. 024 does fall in the rejection region of zlt;-1. 45, therefore I would reject the null and say there is sufficient evidence to indicate mult;50. Interpretation of Results and Conclusion p-value= . 001 .001lt;. 05 Because the p-value of . 001 is less than the significance level of . 05, I will reject the null hypothesis at 5% level. 95% CI=(39. 68, 47. 80)- I am 95% confident that the true mean income lies between $39,680 and $47,800. Minitab Output: One-Sample Z Test of mu = 50 vs lt; 50 The assumed standard deviation = 14. 64 95% Upper N Mean SE Mean Bound Z P 50 43. 74 2. 07 47. 5 -3. 02 0. 001 b. the true population proportion of customers who live in an urban area exceeds 40% 22 people of the 50 surveyed live in an Urban community, which is 44%. My point estimate is . 44. Null and Alternative Hypothesis H0: p=. 40 Ha: pgt;. 40 Level of Significance Level of Significance= . 05 Test Statistic, Critical Value, and Decision Rule Since alpha= . 05, zgt;1. 645, which is upper tailed Rejection region is, zgt;1. 645 To conduct a large sample z-test, I must first determine if the sample size is large enough. nPo= 50(. 40)= 20 and 50(1-. 0)=30 Both are larger than 15, so we conclude that the sample size is large enough to conduct the large sample z test. Z=(. 44-. 40)/. 06928=. 5774. 06928 is calculated by sq-root ((. 4)(. 6))/50)=. 06928 Decision Rule: The calculated test statistic of . 5774 does not fall in the rejection region of zgt;1. 645, theref ore I would not reject H0. There is insufficient evidence to conclude the true population of customers who live in urban communities is greater than 40%. Interpretation of Results and Conclusions p-value= . 282 .282gt;. 05 Because the p-value of . 82 is greater than the significance level of . 05, I will not reject the null. 95% CI= (. 299907, . 587456)- I am 95% confident that the true population proportion of customers who live in an urban area is between 30% and 59%. Minitab Output Test and CI for One Proportion Test of p = 0. 4 vs p gt; 0. 4 95% Lower Sample X N Sample p Bound Z-Value P-Value 1 22 50 0. 440000 0. 324532 0. 58 0. 282 Using the normal approximation. c. the average (mean) number of years lived in the current home is less than 13 years Null and Alternative Hypothesis H0: mu=13 Ha: mult;13 Level of Significance Level of Significance= . 05 Test Statistic, Critical Value, and Decision Rule Since alpha= . 05, zlt;-1. 645, which is lower tailed. Rejection region is zlt;- 1. 645 Calculate the test statistic, x-bar =12. 26 and s=5. 086 Z=(12. 26-13)/. 7193=-1. 03 Decision Rule: The calculated test statistic of -1. 03 does not fall in the rejection region of Zlt;-1. 645, therefore I would not reject the null hypothesis and say there is insufficient evidence to indicate mult;13. Interpretation of Results and Conclusion p-value=. 52 .152gt;. 05 Because the p=value of . 152 is greater than the significance level of . 05, I would not reject the null hypothesis at 5% level. 95% CI=(10. 850, 13. 670)- I am 95% confident that the average number of years lived in current home falls between 10. 85 and 13. 67 years. Minitab Output One-Sample Z Test of mu = 13 vs lt; 13 The assumed standard deviation = 5. 086 95% Upper N Mean SE Mean Bound Z P 50 12. 260 0. 719 13. 443 -1. 03 0. 152 d. the average (mean) credit balance for suburban customers is more than $4300. 5 people of 50 surveyed live in a suburban community, so I will be conducting a t-test because 15lt;30. Null and Alternative Hypothesis H0: mu=$4,300 Ha: mugt;$4,300 Level of Significance Level of Significance= . 05 Test Statistic, Critical Value, and Decision Rule Since alpha= . 05, tgt;1. 761, which is upper tailed. Rejection region is tgt;1. 761 Calculate the test statistic, x-bar=4675 and s=742 T=(4675-4300)/742sqrt15=1. 957 Decision Rule: 1. 957 is greater than 1. 761, which means it does fall in the rejection region, so I would reject H0. Part B Course Project Math 533 Essay Body Paragraphs Because I am rejecting H0, this means that there is sufficient evidence to conclude that the average credit balance for the suburban customers is greater than $4300 Interpretation of Results and Conclusion p-value=. 035 .035lt;. 05 Because the p-value of . 035 is less than the significance level of . 05, I will reject the null hypothesis at 5% level. 95%=(4264, 5086)- I am 95% confident that the average credit balance for suburban customers falls between $4264 and $5,086. Minitab Output One-Sample T Test of mu = 4300 vs gt; 4300 95% Lower N Mean StDev SE Mean Bound T P 15 4675 742 192 4338 1. 96 0. 035 Final Report of Results Before parts a-d are broken down, provided below are a the Seven Elements of a Test Hypothesis that will assist one in understanding the different terminology used in this report. Elements of a Test of Hypothesis 1. Null hypothesis (H0): A theory about the specific values of one or more population parameters. The theory generally represents the status quo, which we adopt until it is proven false. The theory is always stated as H0: parameter = value. 2. Alternative (research) hypothesis (Ha): A theory that contradicts the null hypothesis. The theory generally represents that which we will adopt only when sufficient evidence exists to establish its truth. 3. Test statistic: A sample statistic used to decide whether to reject the null hypothesis. 4. Rejection region: The numerical values of the test statistic for which the null hypothesis will be rejected. The rejection region is chosen so that the probability is? that it will contain the test statistic when the null hypothesis is true, thereby leading to a Type I error. The value of? is usually chosen to be small (e. g. , . 01, . 05, or . 10) and is referred to as the level of significance of the test. 5. Assumptions: Clear statement(s) of any assumptions made about the population(s) being sampled. 6. Experiment and calculation of test statistic: Performance of the sampling experiment and de termination of the numerical value of the test statistic. 7. Conclusion: a. If the numerical value of the test statistic falls in the rejection region, we reject the null hypothesis and conclude that the alternative hypothesis is true. We know that the hypothesis-testing process will lead to this conclusion incorrectly (Type I error) only 100? % of the time when H0 is true. b. If the test statistic does not fall in the rejection region, we do not reject H0. Thus, we reserve judgment about which hypothesis is true. We do not conclude that the null hypothesis is true because we do not (in general) know the probability? that our test procedure will lead to an incorrect acceptance of H0 (Type II error). The average mean annual income was less than $50,000. I rejected the null hypothesis since the p-value of . 01 is smaller than the significance level of . 05. The p-value indicates the probability of rejecting a true null hypothesis. There is a sufficient amount of evidence to support th e claim that the average annual income was less than $50,000 since there is a significance level of . 05. The 95% CI, which is (39. 68, 47. 80) means that I am 95% confident that the true mean (average) income lies between $39,680 and $47,800. Based on these results, the managers speculation that the average annual income was less than $50,000 is correct, which means the null is rejected, and the alternative is accepted. The true population proportion of customers who live in an urban area exceeds 40%. I did not reject the null hypothesis because the p-value of . 282 is larger than the significance level of . 05. The p-value indicates the probability of rejecting a true null hypothesis. There is insufficient evidence to conclude that the true population of customers who live in urban communities is greater than 40%. The 95% CI of (. 299907, . 587456) indicates that we are 95% confident that the true population proportion of customers who live in urban area is between 30% and 59%. Ba sed on these results, the managers speculation that the true population proportion of customers who live in an urban area exceeds 40% is not correct because we decided not to reject the null hypothesis. The average (mean) number of years lived in the current home is less than 13 years. I did not reject the null hypothesis because the p-value of . 152 is greater than the significance level of . 05. The p-value, again, indicates the probability of rejecting a true null hypothesis. There is insufficient evidence to conclude that the average number of years lived in current home is less than 13 years. The 95% CI of (10. 850, 13. 670) indicates that we are 95% confident that the average number of years that customers lived in their current home falls between 11 and 14 years. Based on these results, the managers speculation that the true proportion of customers who have lived in their current home less than 13 years is incorrect because we decided not to reject the null. The average (mean ) credit balance for suburban customers is more than $4300. I did reject the null hypothesis because not only did t (1. 957) fall into the rejection region, but also, the p-value of . 035 is less than . 05. If the p-values is less than the significance level, you will reject the null. There is sufficient evidence to conclude that the average credit balance for the suburban customers is greater than $4300. The 95% confidence intervals indicates that we are 95% confident that the average credit balance for suburban customers falls between $4,264 and $5,086. Based on these results, the managers speculation that the average credit balance for suburban customers is more than $4300 is correct. References Benson, P. G. , McClave, J. T. , amp; Sincich, T. (2011). Statistics for Business and Economics (11th ed. ). Boston, MA: Prentice Hall. We will write a custom essay sample on Part B Course Project Math 533 Essay Example specifically for you for only $16.38 $13.9/page Order now We will write a custom essay sample on Part B Course Project Math 533 Essay Example specifically for you FOR ONLY $16.38 $13.9/page Hire Writer We will write a custom essay sample on Part B Course Project Math 533 Essay Example specifically for you FOR ONLY $16.38 $13.9/page Hire Writer