Question

Please help with all parts of problem 1... problem 2 which is requested for info on problem 1 is also attached

Answer 1

To perform this right-tail hypothesis test, we use the six-step method. Stepl The null and alternative hypotheses are as follows Ho : μ = 475 (The population mean is equal to 475) H. : μ > 475 (The population mean is greater than 475) Step2 We have selected a sample of 100 we decide to use α= 0.05 Step3: B ecause σ is unknown, we us e the 1-d stribution and the-test statistic. Wemust assume that the populati on mean are normally distnbuted. Step4 For a given sample size, n, the test statistic t follows at distribution with-1 degrees of freedom The critical values of the t distributi on with 100-1-99 degrees of freedom The table of t test is obtained by using the excel function =tirv(probabihy.degrees affreedom.) -tinv(0.05,99) = ±1.984217 00991.984217 Step5 The sample mean is 2 -478 The standard deviation is s-100 The sample size is n= 100 The -test statistic is given by 478-475 100 100 10 10 = 0.3 The critical value of t is 00.991.984217 P-value P(t 0.3) -(1-P(t < 0 30)) (1-0.382403) = 0617597 P-value 06176 Step6 Critical Value Approach: Decision Because, the test stati sticist 0.30991.984217 We do not reject the null hypothesis Ho. Since, thepalue 0.6176>a 0.05. We do not reject the null hypothesis Ho he population mea is equal to 475.

To perform this right-tail hypothesis test, we use the six-step method. Stepl The null and alternative hypotheses are as follows Ho : μ = 475 (The population mean is equal to 475) H. : μ > 475 (The population mean is greater than 475) Step2 We have selected a sample of 1000. we decide to use α= 0.05 Step3: B ecause σ is unknown, we us e the 1-d stribution and the-test statistic. Wemust assume that the populati on mean are normally distnbuted. Step4 For a given sample size, n, the test statistic t follows at distribution with-1 degrees of freedom The critical values of the t distributi on with 1000-1- 999 degrees of freedom The table of t test is obtained by using the excel function -tiv(probabihty.deg rees of treedom) i(0.05,999) - 11.64638 Step5 The sample mean is X = 478 The standard deviation is s-100 The sample size is n= 1000 The -test statistic is given by 478-475 1000 31.62278 3 162278 - 0.948683 t0.948683 The critical value of t is 00699.64638 P-value = P(t > 0.948689) 1-P <0.948689)) (1-0.171506) = 0 828494 P-value 08285 Step6 Critical Value Approach: Decision Because, the test stati sticist 0.951.64638 We do not reject the null hypothesis Ho. Since, the p-value-0.8285> 0.05. We do not reject the null hypothesis H0 he population mea is equal to 475.

To perform this right-tail hypothesis test, we use the six-step method. Stepl The null and alternative hypotheses are as follows Ho 475 The popul ati on me an is equal to 475) Ha u>475 (The population mean is greater than 475) Step2 We have selected a sample of 10000 We decide to use a 0.05 Step3: B ecause ơ is unknown, we us e the 1-d stribution and the-test statistic. Wemust assume that the populati on mean are normally distnbuted. Step4 For a given sample size, n, the test statistic t follows az distribution with 2-1 degrees of freedom. The critical values of the t distibuti on with 10000-1- 9999 degrees of freedom The table of t test is obtained by using the excel function =tirv(probabihy.degrees affreedom.) i0.05,9999) = ±1.645006 1.645006 Step5 The sample mean is 2 -478 The standard deviation is s-100 The sample size is-10000 The -test statistic is given by 478-475 10000 100 The critical value of t is1.645006 P-value P(t>3) (1-Pt <3)) (1-0.998647) - 0.001353 P-u00133 Step6 Critical Value Approach: Decision Because, the test stati stistp-1.645006 We reject the null hypothesis Ho Since, the p-value-0.00 1 353 < α-005. We reject the null hypothesis H The population mean is greater than 475

The sample size is n - 100 The sample mean isX 478 The sample standard deviation is s 100 The degrees of freedom is df n-1 =100-1 The level of significance is α-005 The table values are cal culated by using the excel function =tiny(probability, degrees of freedom tinv (0,05,99) 1.984217 -1*1.984217 The formula for calculating the 95% confidence interval for the population mean is 100 100 100 100 100 478-1.984217 μ 478+1.984217 100 478-1.9842171-1 μ 478 +1.984217 478-1.984217(10) μ 478+1.984217(10) 478-19.84216952 μ 478 +19.84216952 458. 1578305,s μ 497.8421695 X+tn-l F-458. 157s& μ 497.8422 Nin Conclusi on We can be 95% confident that the population mean is between 458.1578 and 497.8422

The sample size is n 10000 The sample mean isX 478 The sample standard deviation is s 100 The degrees of freedom is df n-1 -10000-1 The level of significance is α-005 The table values are cal culated by using the excel function tinv(probability, deg rees of jreedom) tin (0.05,9999) 1.960201 N-1*1.960201 The formula for calculating the 95% confidence interval for the population mean is 100 10000 100 10000 478-1.960201 478+1.960201 100 100 μ (1 )s 100 100 ( 0) 478-1.960201 μ 478 + 1.960201 478-1.960201(1)Su s478+1.960201(1) 478-1 .960201 μ 478 +1.960201 476.0398SS479.9602 -i F-476, 039s&AS 479.9602 Nin Conclusi on We can be 95% confident that the population mean is between 476.0398 and 479.9602