**Hypothesis Test for the Difference of Population Means: Paired**(a) **State the Null and Alternative Hypothesis:** - \( H_0 \): [ ] - \( H_1 \): [ ](b) **Determine the Type of Test Statistic to Use:** - Type of test statistic: (Choose one) [Dropdown menu](c) **Find the Value of the Test Statistic:** - [Input box] (Round to three or more decimal places.)(d) **Find the Critical Value at the 0.05 Level of Significance:** - [Input box] (Round to three or more decimal places.)(e) **Conclusion at the 0.05 Level:** - Can the company conclude that the mean assembly time for Process 1 exceeds that of Process 2? - [ ] Yes - [ ] No A computer manufacturer is interested in comparing assembly times for two keyboard assembly processes. Process 1 is the standard process used for several years, and Process 2 is an updated process hoped to bring a decrease in assembly time. Assembly times can vary considerably from worker to worker, and the company decides to eliminate this effect by selecting 8 workers at random and timing each worker on each assembly process. Half of the workers are chosen at random to use Process 1 first, and the rest use Process 2 first. For each worker and each process, the assembly time (in minutes) is recorded, as shown in the table below:| Worker (k) | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 ||------------|----|----|----|----|----|----|----|----|| Process 1 | 79 | 39 | 56 | 72 | 49 | 90 | 75 | 67 || Process 2 | 74 | 19 | 60 | 77 | 38 | 75 | 44 | 44 || Difference (Process 1 - Process 2) | 5 | 20 | -4 | -5 | 11 | 15 | 31 | 23 |Based on these data, can the company conclude, at the 0.05 level of significance, that the mean assembly time for Process 1 exceeds that of Process 2? Answer this question by performing a hypothesis test regarding \(\mu_d\) (which is \(\mu\) with a letter "d" subscript), the population mean difference in assembly times for the two processes. Assume that this population of differences (Process 1 minus Process 2) is normally distributed.Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places and round your answers as specified. (If necessary, consult a list of formulas.)(a) State the null hypothesis \(H_0\) and the alternative hypothesis \(H_1\).- \(H_0\): - \(H_1\): Carry out the hypothesis test using these hypotheses to determine if Process 1's mean assembly time significantly exceeds that of Process 2.

Given, Worker Process 1 Process 2 difference d = (Process 2-Process 1) 1 79 74 -5 2 39 19…

Answered: (a) State the null hypothesis Ho and…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Find the p-value for a left-tailed hypothesis test with a test statistic of z= -1.25
Q3 Test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, and then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. Among 2125 passengers cars in a particular region, 229 had only rear license plates. Among 326 commercial trucks, 47 had only rear license plates. A reasonable hypothesis is that commercial trucks owners violate laws requiring front license plates at a higher rate than owners of passengers cars. Use a 0.01 significance level to test that hypothesis. A. Identity the nulls and alternative hypotheses B. Identify the test statistic ( round to two decimal places as needed) C. Identify the p- values (round to two decimal places as needed ? State the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. A. Fail to reject Ho. There is not sufficient evidence to support the claim that commercial trucks owners…
The P-value for a hypothesis test is shown. Use the P-value to decide whether to reject Ho when the level of significance is (a) a = 0.01, (b) a = 0.05, and (c) a = 0.10. P= 0.0961 (a) Do you reject or fail to reject Ho at the 0.01 level of significance? O A. Reject Ho because the P-value, 0.0961, is greater than a =0.01. O B. Fail to reject Ho because the P-value, 0.0961, is greater than a = 0.01. O C. Reject Ho because the P-value, 0.0961, is less than a = 0.01. O D. Fail to reject Ho because the P-value, 0.0961, is less than a = 0.01. (b) Do you reject or fail to reject Ho at the 0.05 level of significance? O A. Fail to reject Ho because the P-value, 0.0961, is less than a = 0.05. O B. Fail to reject Ho because the P-value, 0.0961, is greater than a = 0.05. O C. Reject Ho because the P-value, 0.0961, is less than a = 0.05. O D. Reject Ho because the P-value, 0.0961, is greater than a = 0.05. (c) Do you reject or fail to reject Ho at the 0.10 level of significance? O A. Fail to…
The P-value for a hypothesis test is shown. Use the P-value to decide whether to reject Ho when the level of significance is (a) a = 0.01, (b) a =0.05, and (c) a= 0.10. P= 0.1089 (a) Do you reject or fail to reject Ho at the 0.01 level of significance? O A. Reject Ho because the P-value, 0.1089, is greater than a= 0.01. O B. Reject H, because the P-value, 0.1089, is less than a= 0.01. O C. Fail to reject Ho because the P-value, 0.1089, is less than a = 0.01. D. Fail to reject Ho because the P-value, 0.1089, is greater than a = 0.01. (b) Do you reject or fail to reject Ho at the 0.05 level of significance? O A. Fail to reject Ho because the P-value, 0.1089, is greater than a = 0.05. 0.05. O B. Reject H, because the P-value, 0.1089, is greater than O C. Fail to reject Ho because the P-value, 0.1089, is less than a = 0.05. O D. Reject H, because the P-value, 0.1089, is less than a = 0.05.
An institute conducted a clinical trial of its methods for gender selection. The results showed that 153 of 276 babies born to parents using a specific gender-selection method were boys. Use the sign test and a 0.05 significance level to test the claim that the method has no effect on the likelihood of a boy. Let p denote the population proportion of baby boys. Ignore the possibility of the method decreasing the likelihood of a boy. What are the null and alternative hypotheses? O A. Ho: p=0.5 H₂₁: p20.5 O C. Ho: p=0.5 H₁: p=0.5 O B. Ho:p>0.5 H₁: p=0.5 O D. Ho: p=0.5 H₁: p > 0.5 Find the test statistic. Test statistic = Find the P-value. P-value = (Round to four decimal places as needed.) Determine the proper conclusion. Choose the correct answers below. Since the P-value is (Round to two decimal places as needed.) than the significance level, evidence that the method used is effective in increasing the likelihood of a boy. the null hypothesis. There is
Test the claim that the mean GPA of night students is larger than 2.4 at the 0.025 significance level. The null and alternative hypothesis would be: Ho: µ = — 2.4 Но:р 2.4 Н: + 2.4 H:p> 0.6 H:д> 2.4 Hi:p #0.6 Hi:p < 0.6 Н:р < 2.4 The test is: left-tailed two-tailed right-tailed Based on a sample of 65 people, the sample mean GPA was 2.41 with a standard deviation of 0.03 The test statistic is: (to 2 decimals) The p-value is: (to 2 decimals) Based on this we: O Reject the null hypothesis O Fail to reject the null hypothesis
Consider the following hypothesis test. Ho: H₁ - H₂ SO Ha: M₁ - H₂ > O The following results are for two independent samples taken from the two populations. Sample 1 Sample 2 n₁ = 40 7₂ = 50 X₁ = 25.4 %1 = 5.7 (a) What is the value of the test statistic? (Round your answer to two decimal places.) = 22.8 x₂ = %₂ = 6 (b) What is the p-value? (Round your answer to four decimal places.) (c) with a = 0.05, what is your hypothesis testing conclusion? Reject Ho. There is insufficient evidence to conclude that μ₁ - H₂ > 0. Reject Ho. There is sufficient evidence to conclude that μ₁ - μ₂ > 0. Do not reject Ho. There is insufficient evidence to conclude that μ₁ - μ₂ > 0. Do not Reject Ho. There is sufficient evidence to conclude that μ₁ −μ₂ > 0. -
The P-value for a hypothesis test is shown. Use the P-value to decide whether to reject Ho when the level of significance is (a) a= 0.01, (b) a=0.05, and (c) a= 0.10. P=0.0596 (a) Do you reject or fail to reject Ho at the 0.01 level of significance? OA Fail to reject Hg because the P-value, 0.0596, is greater than a= 0.01. OB. Reject Ho because the P-value, 0.0596, is greater than a= 0.01. OC. Fail to reject Hg because the P-value, 0.0596, is less than a=0.01. %3D OD. Reject Ho because the P-value, 0.0596, is less than a= 0.01. %3D
Test the claim that the proportion of men who own cats is larger than 20% at the .05 significance level. The null and alternative hypothesis would be: 0.2 Но:р — 0.2 Но: и 0.2 Ho:H Ho:H H:р + 0.2 Н]:р 0.2 Hi:р > 0.2 Hi:р#0.2 0.2 Но:р || The test is: left-tailed two-tailed right-tailed Based on a sample of 25 people, 25% owned cats The test statistic is: (to 2 decimals) The critical value is: (to 2 decimals) Based on this we: O Fail to reject the null hypothesis O Reject the null hypothesis
Machines at a factory are supposed to dispense 250 grams of sugar per pack. A random sample of sugar packs is taken from the factory, and their contents (in grams) are measured. Following are the sample data: 247 247 247 248 249 250 250 250 251 251 Is there sufficient evidence to conclude that the machines do not conform to the indicated sugar amount? Use 0.10 level of significance. Which case for hypothesis testing must be used? ( Select ] What is the alternative hypothesis? ( Select ] What is the value of the test statistic? [ Select ] What is/are the critical value/s? Select ] What is the appropriate conclusion? [Select
Three students, an athlete, a fraternity member, and an honors student, record the number of hours they slept each night for 20 nights. O JMP Applet imp ? Oneway Analysis of Sleep Hours By Student Oneway Analysis of Sleep Hours By Student 10 Oneway Anova 14 Summary of Fit 12 Rsquare 0.024506 10- Adj Rsquare Root Mean Square Error Mean of Response Observations (or Sum Wgts) -0.00072 1.99517 7.7 60 Analysis of Variance Sum of Mean F Prob > Source DF Squares Square Ratio F 2 Athiete Frat Honors Student 2 5.70000 2.85000 0.7180 0.4931 Student Error 57 226.90000 3.98070 C. Total 59 232.60000 Oneway Anova Means for Oneway Anova Std Lower Upper 95% Summary of Fit Level Number Mean Error 95% Athlete 20 8.10000 0.44813 7.2086 8.9934 Rsquare 0.024506 Frat 20 7.65000 0.44813 6.7588 8.5434 Adj Rsquare Root Mean Square Error Mean of Response Observations (or Sum Wgts) -0.00972 Honors 20 7.35000 0.44813 6.4586 8.2434 1.99517 Std Error uses a pooled estimate of error variance 7.7 60 Analysis of…

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(a) State the null hypothesis Ho and the alternative hypothesis H₁. H:0 H₁:0 b) Determine the type of test statistic to use,