b. The true population proportion of customers who live in an urban area exceeds 40% c. The average (mean) number of years lived in the current home is less than 13 years d. The average (mean) credit balance for suburban customers is more than $4300 I have been assigned to analyze the speculated data listed above by performing hypothesis test for each of the above situations (using the Seven elements of a Test Hypothesis with a=.05) in order to see if there is evidence to support my manager’s beliefs in each case (a-d), explain my conclusion in simple terms, compute the p-value with the interpretation, follow up with computing 95% confidence intervals for each of the variables described in a. b. c. d. along with interpreting these intervals. This paper will also include an Appendix with all the steps in hypothesis testing, as well as the confidence intervals and Minitab output In order to understand how hypothesis testing is done it is important that you know the elements of the Test of Hypothesis, and what each step means. The Seven elements of a Test of Hypothesis are: 1. Null
For what values of t will the null hypothesis not be rejected? a) To the left of -1.645 or to the right of 1.645 b) To the left of -1.345 or to the right of 1.345 c) Between -1.761 and 1.761 d) To the left of -1.282 or to the right of 1.282 2. Which of the following is a characteristic of the F distribution? a) Normally distributed b) Negatively skewed c) Equal to the t-distribution d) Positively skewed Complete Answers here QNT 561 Final Exam 3. For a chi-square test involving a contingency table, suppose the null hypothesis is rejected.
In short the majority of AJ Davis customers are Urban dwellers. In addition a Pie-Chart below depicts the same data. 2) Income Based on the histogram above which used a sample of 50 customers from AJ Davis store the mean income is about $45.650 and it follow a normal curve or has what we call the bell curve. Descriptive Statistics: household size Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3 household size 51 0 4.431 0.357 2.548 1.000 2.000 4.000 7.000 Variable Maximum household size 8.000 From a sample of 50 customers the average household is 4.4 and also the median household is 4. The maximum household is 8 and the minimum is 1.
Rochelle decided to compare two plants for 50 days, Memphis plant-which will have the vector drive and Birmingham plant-which will use the existing system and prepare a report. The financial analysts believe that the purchase can be justified if the equipment leads to the average increase in production of atleast 10,000 bricks per day. Since we have to check whether the difference between the mean value of the bricks per day at Memphis plant and Birmingham plant is greater than 10000(for purchase to be justified), we will use hypothesis testing for means and compare the sampling distribution of the mean value of bricks produced in a day. This method is the typical method to solve these kind of problems. Data available is Plant Total Bricks produced in 50 days S (standard deviation) x̄ (mean) Memphis 7484500 3402.46 149690 Birmingham 6902350 3364.68 138047 Hypothesis test In a hypothesis test we assume Ho to be true and try to find evidence that shows otherwise.
If you choose 40 random employees from the corporation, the standard error would equal 6/Square root of 40 = .95 days. The 12 days in this department corresponds to (12-8.2)/.95 = 4 standard errors above the corporation average of 8.2. This is much higher than two or three standard errors, and it appears to be beyond chance variation. Chapter 9 Exercise 3 The p- value tells you how likely it would be to get results at least as extreme as this if there was no difference in the taste and only chance variation was operating. In this problem, p-value of 0.02 means that, if there is no difference in taste, then there is only 2% chance that 70% or more people would declare one drink better than the
For what values of t will the null hypothesis not be rejected? a) To the left of -1.645 or to the right of 1.645 b) To the left of -1.345 or to the right of 1.345 c) Between -1.761 and 1.761 d) To the left of -1.282 or to the right of 1.282 QNT 561 Final Questions and Answers QNT 561 Final Exam 2. Which of the following is a characteristic of the F distribution? a) Normally distributed b) Negatively skewed c) Equal to the t-distribution d) Positively skewed 3. For a chi-square test involving a contingency table, suppose the null hypothesis is rejected.
To better understand the size of a household for AJ Davis’s credit customers we used a table of descriptive statistics, frequency table, and a bar chart. Descriptive Statistics: Size Variable Mean SE Mean StDev Variance Minimum Q1 Median Q3 Size 3.420 0.246 1.739 3.024 1.000 2.000 3.000 5.000 N for Variable Maximum Mode Mode Size 7.000 2 15 Tally for Discrete Variables: Size Size Count CumCnt 1 5 5 2 15 20 3 8 28 4 9 37 5 5 42 6 5 47 7 3 50 N= 50 This information tells us that the mean, or average, household size of the credit customer is 3.42. The median, or middle, household size is 3. There is a standard deviation of 1.739. As can be seen in the bar chart and the frequency table, the mode, or most frequent occurring household size, is 2 with 15 out of the 50 households being this size.
1.328 b. 2.539 c. 1.325 d. 2.528 ANSWER: a -go to the t-dtistrubution and use α=0.20 or confidence of 80% and use dof=19 3. Read the t statistic from the table of t distributions and circle the correct answer. A one-tailed test (upper tail), a sample size of 18 at a .05 level of significance t = a. 2.12 b.
TOPIC 8 Chi-Square goodness-of-fit test Problem 12.1 Use a chi-square goodness-of-fit to determine whether the observed frequencies are distributed the same as the expected frequencies (α = .05) Category | fo | fe | 1 | 53 | 68 | 2 | 37 | 42 | 3 | 32 | 33 | 4 | 28 | 22 | 5 | 18 | 10 | 6 | 15 | 8 | Step 1 Ho: The observed frequencies are distributed the same as the expected frequencies Ha: The observed frequencies are not distributed the same as the expected frequencies Step 2 df = k – m – 1 Step 3 α = 0.05 x 2 0.05, 5df = 11.0705 Step 4 Reject Ho if x 2 > 11.0705 Category | fo | fe | | 1 | 53 | 68 | | 2 | 37 | 42 | | 3 | 32 | 33 | | 4 | 28 | 22 | | 5 | 18 | 10 | | 6 | 15 | 8
February 2005, reports on the results achieved by Bank of America in improving customer satisfaction and customer loyalty by listening to the ‘voice of the customer.’ A key measure of customer satisfaction is the response on a scale from 1 to 10 to the question: “Considering all the business you do with Bank of America, what is your overall satisfaction with Bank of America?” Suppose that a random sample of 350 current customers results in 195 customers with a response of 9 or 10 representing “customer delight.” Find a 95 percent confidence interval for the true proportion of all current Bank of America customers who would respond with a 9 or 10. Are we 95 percent confident that this proportion exceeds .48, the historical proportion of customer delight for Bank of