**Educational Content on Hypothesis Testing and Confidence Intervals****Independent Random Samples Analysis**Given are two independent random samples from normal populations. The results are depicted in a table as follows:| Sample 1 | Sample 2 ||----------|----------|| 1.5 | 3.4 || 2.6 | 2.9 || 2.8 | 3.9 || 1.7 | 3.1 || 1.5 | |Assume equal variances for the analysis.**Statistical Analysis Tasks**1. **t-Statistic Calculation** - Calculate the t-statistic and round to two decimal places as needed.2. **Observed Significance Level** - Determine the p-value and round to three decimal places as needed. 3. **Results Interpretation** - Choose the correct interpretation: - A. Do not reject \( H_0 \). There is not sufficient evidence to indicate that \( \mu_2 > \mu_1 \). - B. Reject \( H_0 \). There is not sufficient evidence to indicate that \( \mu_2 > \mu_1 \). - C. Reject \( H_0 \). There is sufficient evidence to indicate that \( \mu_2 > \mu_1 \). - D. Do not reject \( H_0 \). There is sufficient evidence to indicate that \( \mu_2 > \mu_1 \).4. **Confidence Interval Computation** - Calculate a 90% confidence interval for \( (\mu_1 - \mu_2) \). - Provide the confidence interval and round to two decimal places as needed.5. **Inferential Procedures Evaluation** - Determine which gives more information about \( (\mu_1 - \mu_2) \): - A. The hypothesis test provides more information about \( (\mu_1 - \mu_2) \) because it gives a range estimate of \( \mu_1 \) and \( \mu_2 \), while the confidence interval only addresses whether \( \mu_2 > \mu_1 \). - B. The hypothesis test provides more information about \( (\mu_1 - \mu_2) \) because it gives a ranged estimate of \( \mu_1 - \mu_2 \), while the **Transcription for Educational Website**---**Independent random samples from normal populations produced the results shown in the table to the right. Complete parts a through d below. Assuming Equal Variances**| Sample 1 | Sample 2 ||----------|----------|| 1.5 | 3.4 || 2.6 | 2.9 || 2.8 | 3.9 || 1.7 | 3.1 || 1.5 | |---**a. Calculate the pooled estimate of \(\sigma^2\).**\[ s_p^2 = \, [ \text{Box for input} ] \](Round to four decimal places as needed.)---**b. Do the data provide sufficient evidence to indicate that \(\mu_2 > \mu_1\)? Test using \(\alpha = 0.05\).****What are the null and alternative hypotheses?**- ⭘ A. \( H_0: \mu_1 - \mu_2 \geq 0 \) \( H_a: \mu_1 - \mu_2 < 0 \)- ⭘ B. \( H_0: \mu_1 - \mu_2 \leq 0 \) \( H_a: \mu_1 - \mu_2 > 0 \)- ⭘ C. \( H_0: \mu_1 - \mu_2 = 0.05 \) \( H_a: \mu_1 - \mu_2 \neq 0.05 \)- ⭘ D. \( H_0: \mu_1 - \mu_2 \geq 0.05 \) \( H_a: \mu_1 - \mu_2 \leq 0.05 \)**What is the test statistic?**\[ t = \, [ \text{Box for input} ] \](Round to two decimal places as needed.)---**What is the observed significance level?**\[ \text{p-value} = \, [ \text{Box for input} ] \](Round to three decimal places as needed.)---**Interpret the results. Choose the correct answer below.**- ⭘ A. Do not reject \(H

Answered: ..... (Round to two decimal places as…

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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Please use the chart to answer the questions on the second image (all of which I got wrong the first time) this is not graded, it is a homework practice question.
The claim is that smokers have a mean cotinine level greater than the level of 2.84 ng/mL found for nonsmokers. (Cotinine is used as a biomarker for exposure to nicotine.) The sample size is n=761 and the test statistic is t=55.668. Use technology to find the P-value. Based on the result, what is the final conclusion? Use a significance level of 0.10. State the null and alternative hypotheses. H₁> 2.84 2.84 (Type integers or decimals. Do not round.) The test statistic is (Round to two decimal places as needed.)
Assume that the significance level is a = 0.01. Use the given statement to find the critical value(s). The test statistic of z = - 1.66 is obtained when testing the claim that p =: Click here to view page 1 of the Normal table. Click here to view page 2 of the Normal table. The critical value(s) is/are. (Round to three decimal places as needed. Use a comma to separate answers as needed.)
Consider a drug that is used to help prevent blood clots in certain patients. In clinical trials, among 6052 patients treated with this drug, 149 developed the adverse reaction of nausea. Use a 0.05 significance level to test the claim that 3% of users develop nausea. Does nausea appear to be a problematic adverse reaction? Identify the null and alternative hypotheses for this test. Choose the correct answer below. O A. Ho: p=0.03 H;:p>0.03 O B. Ho: p#0.03 H;:p= 0.03 OC. Ho: p= 0.03 H4: p+0.03 O D. Ho: p= 0.03 H:p<0.03 Identify the test statistic for this hypothesis test. The test statistic for this hypothesis test is (Round to two decimal places as needed.) Identify the P-value for this hypothesis test. The P-value for this hypothesis test is. (Round to three decimal places as needed.) Identify the conclusion for this hypothesis test.
You wish to test the following claim at a significance level of a=0.05. H0:P=0.19 H1:P<0.19You obtain a sample of size 739 in which there are 119 successful observations.What is the test statistic for this sample? (Report answer accurate to 2 decimal places but do not round in the middle of your calculations.)What is the p-value for this sample? (Report answer accurate to 4 decimal places.)This test statistic leads to a decision to reject the null accept the null fail to reject the null As such, the final conclusion is that there is sufficient evidence to conclude that the population proportion is less than 0.19. there is not sufficient evidence to conclude that the population proportion is less than 0.19. there is sufficient evidence to conclude that the population proportion is equal to 0.19. there is not sufficient evidence to conclude that the population proportion is equal to 0.19.
Need help with 7 a and b.
Consider a drug that is used to help prevent blood clots in certain patients. In clinical trials, among 5973 patients treated with this drug, 153 developed the adverse reaction of nausea. Use a 0.10 significance level to test the claim that 3% of users develop nausea. Does nausea appear to be a problematic adverse reaction? Identify the null and alternative hypotheses for this test. Choose the correct answer below. O A. Ho:p= 0.03 H:p0.03 Identify the test statistic for this hypothesis test. The test statistic for this hypothesis test is (Round to two decimal places as needed.)
. What is the appropriate analysis? Using statistical equations, give complete statements of the null and alternative hypotheses. Give any appropriate equations that may be generated by the analysis.
What does brand loyalty mean to consumers? According to a recent research report, 38% of consumers associate trust with brand loyalty. Complete parts (a) through (e) below. To conduct a follow-up study that would provide 99% confidence that the point estimate is correct to within ±0.04 of the population proportion, how many people need to be sampled? A sample size of __ consumers is needed.
Assume a significance level of a 0.05 and use the given information to complete parts (a) and (b) below Original claim: The proportion of male golfers is more than 0.9 The hypothesis test results in a P-value of 0 056. a. State a conclusion about the null hypothesis. (Reject H, or fail to reject H, ) Choose the corect answer below. O A. Fal to reject H, because the P-value is less than a OB. Fal to reject H, because the P-value in greater than it OC. Reject H, because the P-value is less than a OD. Reject Hg because the P-value is greater than
You wish to test the following claim at a significance level of a = 0.01. Ho:p = 0.68 H1:p < 0.68 You obtain a sample of size 547 in which there are 343 successful observations. What is the test statistic for this sample? (Report answer accurate to 2 decimal places but do not round in the middle of your calculations.) What is the p-value for this sample? (Report answer accurate to 4 decimal places.) This test statistic leads to a decision to reject the null accept the null U fail to reject the null As such, the final conclusion is that there is sufficient evidence to conclude that the population proportion is less than 0.68. there is not sufficient evidence to conclude that the population proportion is less than 0.68. there is sufficient evidence to conclude that the population proportion is equal to 0.68. there is not sufficient evidence to conclude that the population pronortion is ogu

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