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I need help with hypothesis testing and excel. I have attached the sample database for use. Can you please explain how this is done in excel and what the statistical formulas are. I do know that for a hypothesis test with .05 or 95% must compute the z-value which is actual - predit/som var - how do I do this. 1. Test a hypothesis to see whether the average overall job satisfaction (in the population of all workers in the USA) is equal to 4.5 with a = .05. a. State the null hypothesis, the alternative hypothesis, and the significance level. Null hypothesis H0: the average overall job satisfaction (in the population of all workers in the USA) is equal to 4.5 Alternative hypothesis H1: the average overall job satisfaction (in the population of all workers in the USA) is NOT equal to 4.5 The significance level is alpha=0.05 b. Using the data in our database, calculate the test statistic. Denote the sample mean by x and the sample standard deviation by s, and sample size n Use excel, we get Column1 Mean Standard Error Median Mode Standard Deviation Sample Variance Kurtosis Skewness Range Minimum Maximum Sum Count 4.330208 0.080387 4.4 4.6 1.364211 1.86107 -0.46875 -0.22289 6 1 7 1247.1 288 So, x =4.330208, s=1.364211 and n=288. So, we can compute the t-statistic t x 4.5 s n 4.3302084.5 1.364211 288 2.112 c. What is the critical level for the significance level? Degree of freedom is n-1=288-1=287. Use a website http://duke.usask.ca/~rbaker/Tables.html , we get the critical level for the significance level is 1.968. d. What is your conclusion? Do we accept or reject the null hypothesis? Since |t|=2.112>1.968, we should reject the null hypothesis H0. So, we conclude that the average overall job satisfaction (in the population of all workers in the USA) is NOT equal to 4.5 You may use Excel for the calculations, but you need to answer all four parts of this question. 2. Propose a hypothesis test for the mean intrinsic job satisfaction, similar to the test from problem 1, and answer parts a, b, c, and d of problem 1 for this hypothesis test. You may use Excel for the calculations, but you need to answer the four questions. a. State the null hypothesis, the alternative hypothesis, and the significance level. Null hypothesis H0: the average intrinsic job satisfaction (in the population of all workers in the USA) is equal to 4.5 Alternative hypothesis H1: the average intrinsic job satisfaction (in the population of all workers in the USA) is NOT equal to 4.5 The significance level is alpha=0.05 b. Using the data in our database, calculate the test statistic. Denote the sample mean by x and the sample standard deviation by s, and sample size n Use excel, we get Column1 Mean Standard Error Median Mode Standard Deviation Sample Variance Kurtosis Skewness Range Minimum Maximum 5.219722 0.058936 5.17 5 1.000183 1.000366 0.870918 -0.4306 6 1 7 Sum Count 1503.28 288 So, x =5.219722, s=1.000183 and n=288. So, we can compute the t-statistic t x 4.5 s n 5.2197224.5 1.000183 288 12.21 c. What is the critical level for the significance level? Degree of freedom is n-1=288-1=287. Use a website http://duke.usask.ca/~rbaker/Tables.html , we get the critical level for the significance level is 1.968. d. What is your conclusion? Do we accept or reject the null hypothesis? Since |t|=12.21>1.968, we should reject the null hypothesis H0. So, we conclude that the average intrinsic job satisfaction (in the population of all workers in the USA) is NOT equal to 4.5. 3. We believe that half of the population would have an extrinsic job satisfaction of 5.0 or greater. Answer parts a, b, c, and d of problem 1 for this hypothesis test of a proportion. You may use Excel for the calculations, but you need to answer the four questions. a. State the null hypothesis, the alternative hypothesis, and the significance level. Null hypothesis H0: the average extrinsic job satisfaction (in the population of all workers in the USA) is equal to 5.0 Alternative hypothesis H1: the average extrinsic job satisfaction (in the population of all workers in the USA) is greater than 5.0 The significance level is alpha=0.05 b. Using the data in our database, calculate the test statistic. Denote the sample mean by x and the sample standard deviation by s, and sample size n Use excel, we get Column1 Mean Standard Error Median Mode Standard Deviation Sample Variance 4.911111 0.096975 5.5 6 1.645715 2.708378 Kurtosis Skewness Range Minimum Maximum Sum Count -0.16404 -0.9033 6 1 7 1414.4 288 So, x =4.911111, s=1.645715 and n=288. So, we can compute the t-statistic t x 4.5 s n 4.9111115 1.645715 288 1.031 c. What is the critical level for the significance level? Degree of freedom is n-1=288-1=287. Use a website http://duke.usask.ca/~rbaker/Tables.html , we get the critical level for the significance level is 1.65. d. What is your conclusion? Do we accept or reject the null hypothesis? Since |t|=1.031<1.65, we fail to reject the null hypothesis H0. So, we conclude that the average extrinsic job satisfaction (in the population of all workers in the USA) is greater than 5.0. 4. We believe that the variance of the overall job satisfaction is equal to 1.0 Answer parts a, b, c, and d of problem 1 for this hypothesis test of a variance. You may use Excel for the calculations, but you need to answer the four questions. a. State the null hypothesis, the alternative hypothesis, and the significance level. Null hypothesis H0: The variance of the overall job satisfaction is equal to 1.0 Alternative hypothesis H1: The variance of the overall job satisfaction is NOT equal to 1.0 The significance level is alpha=0.05 b. Using the data in our database, calculate the test statistic. Denote the sample mean by x and the sample standard deviation by s, and sample size n Use excel, we get Column1 Mean Standard Error Median Mode Standard Deviation 4.330208 0.080387 4.4 4.6 1.364211 Sample Variance Kurtosis Skewness Range Minimum Maximum Sum Count 1.86107 -0.46875 -0.22289 6 1 7 1247.1 288 So, s=1.364211 and n=288. So, we can compute the chi-squared statistic 2 ( n1) s 287 *1.3642112 534.12 2 2 c. What is the critical level for the significance level? Degree of freedom is n-1=288-1=287. Use a website http://duke.usask.ca/~rbaker/Tables.html , we get the critical level for the significance level is 327.512 d. What is your conclusion? Do we accept or reject the null hypothesis? Since 2 534.12 327.512 , we should reject the null hypothesis H0. So, we conclude that the variance of the overall job satisfaction is NOT equal to 1.0. 5. We will call a “deskbody” a person whose intrinsic job satisfaction level is higher than their extrinsic job satisfaction level (i.e. happy with their job more than their office). We will call a “socialbody” a person whose extrinsic job satisfaction level is higher than their intrinsic job satisfaction level (i.e. happy with the office more than their job). We believe that there are equal deskbodies and socialbodies in the work force. a. State an appropriate null hypothesis and its alternative hypothesis. Null hypothesis H0: there are equal deskbodies and socialbodies in the work force Alternative hypothesis H1: there are not equal deskbodies and socialbodies in the work force. b. In our database, what percent of the employees are deskbodies? Are socialbodies? From given information, we count that there are 141 deskbodies and 140 socialbodies. So, 141/288=48.96% of the employees are deskbodies, and 140/288=48.61% of the employees are socialbodies. c. What did you do with the employees who had equal intrinsics and extrinsics? When we test the null hypothesis H0, we can remove those people who have equal intrinsics and extrinsics. 6. Determine the required sample size if you need to estimate the number of workers in the United States who are highly satisfied with their job and you want the estimate to be within 2 percentage points with a 96% confidence interval. If we choose the overall job satisfaction, then we use Excel to get Column1 Mean Standard Error Median Mode Standard Deviation Sample Variance Kurtosis Skewness Range Minimum Maximum Sum Count 4.330208 0.080387 4.4 4.6 1.364211 1.86107 -0.46875 -0.22289 6 1 7 1247.1 288 The sample standard deviation is s=1.364211. So, t-value=2.05 if we use 96% significance level. So, the standard error E s t . By hypothesis, E<=2%. So, the required sample size n n can be determined by E s n t 2% So, 1.364211 n * 2.05 2% So, the standard error n ( 1.364211 * 2.05) 2 19552.89 2% So, we choose the minimum sample size n=19553 So, the required sample size is 19553 if you need to estimate the number of workers in the United States who are highly satisfied with their job and you want the estimate to be within 2 percentage points with a 96% confidence interval. hypothesis testing ------------------------------------------------------------------------------I need help with hypothesis testing and excel. I have attached the sample database for use. Can you please explain how this is done in excel and what the statistical formulas are. When we do hypothesis testing, we need to know what the sample mean and sample standard deviation are. We can get them by using Excel. For example, in question 1), we use Excel to get Column1 Mean Standard Error Median Mode Standard Deviation Sample Variance Kurtosis Skewness Range Minimum Maximum Sum Count 4.330208 0.080387 4.4 4.6 1.364211 1.86107 -0.46875 -0.22289 6 1 7 1247.1 288 How can you get it by Excel? Open your excel file, choose “Tools” on the top menu, choose “Data Analysis” and click it, you will have a window from which you choose “Descriptive Statistics” and click “OK”. You will have another window in which you need to input $F$2:$F$289 if you want to get something about the Overall satisfaction—column F; if you want to get something about “Intrinsic”column G, then you need to input $G$2:$G$289; then click” Descriptive Statistics” and “OK”, you will get a table above. I do know that for a hypothesis test with .05 or 95% must compute the z-value which is actual - predit/som var - how do I do this. In many times, you have to compute the t-value since we don’t use Ztest but t-test. If we know the standard deviation of population, then we can use Z-test. In this case, we need to compute z-value. If we don’t know the standard deviation of population, then we can use ttest. In this case, we need to compute t-value. Either case, we can use a website http://duke.usask.ca/~rbaker/Tables.html (maybe you can look up some distribution tables or critical values tables) to get zvalue or t-value. If you use this website, you need to choose “critical values for z-test” or “critical values for t-test” and click ”GO”. For example, if you choose “critical values for z-test”, then you need to select “1-tailed” or “two-tailed” and “significance level”, for instance, alpha=0.05—you can select 0.05 from a list. If alpha=0.01, then you select 0.01 from a list. Then you will get the critical value, namely, zvalue. If alpha=0.05, we can get z-value=1.645 if it is a “1-tailed” test; z-value=1.96 if it is a “two-tailed” test. Hope it makes sense. If you have further questions, feel free to ask me. Good luck! Changping