a. Calculate the Standard Error oz-in (3 decimal places) b. Calculate the Z-test statistic using the standard error from part a. z = (2 decimal places) C. At a = 0.1, Use the distribution table to find the critical values for the rejection region %3D (3 decimal places)
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Given : n1=41 , n2=42 , X1-bar=42.69 , X2-bar=42.43 , σ1=0.57 , σ2=0.53 , α=0.1
Our aim is to find the following :
a) Standard error
b) Z-test statistic
c) Critical value
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- One group of accounting students took a distance learning class, while another group took the same course in a traditional classroom. At α = .10, is there a significant difference in the mean scores listed below? Exam Scores for Accounting Students Statistic Distance Classroom Mean scores x¯1x¯1 = 9.7 x¯2x¯2 = 10.0 Sample std. dev. s1 = 2.7 s2 = 2.9 Number of students n1 = 23 n2 = 23 Specify the decision rule. (Round your answers to 3 decimal places. A negative value should be indicated by a minus sign.) Find the test statistic tcalc. (Round your answer to 4 decimal places. A negative value should be indicated by a minus sign.) Use Excel to find the p-value. (Round your answer to 3 decimal places.)Data on the weights (lb) of the contents of cans of diet soda versus the contents of cans of the regular version of the soda is summarized to the right. Assume that the two samples are independent simple random samples selected from normally distributedpopulations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b) below. Use a 0.01 significance level for both parts. Diet Regular μ μ1 μ2 n 33 33 x 0.78488 lb 0.80356 lb s 0.00444 lb 0.00744 lb a. Test the claim that the contents of cans of diet soda have weights with a mean that is less than the mean for the regular soda. i) What are the null alternative hypothesis? ii) What's the test statics? iii) What's the P-value? iv) State the conclusion vi) construct a confidence interval suitable suitable for testing the claim that the two samples are from populations with the same mean vii) Does the confidence interval…Find the critical value(s) and rejection region(s) for a left-tailed chi-square test with a sample size n=22 and level of significance α=0.05. LOADING... Click the icon to view the Chi-Square Distribution Table. Find the critical value(s). enter your response here (Round to three decimal places as needed. Use a comma to separate answers as needed.) Find the Rejection Region(s).
- The corrosive effects of various soils on coated and uncoated steel pipe was tested by using a dependent sampling plan. The data collected are summarized below, where d is the amount of corrosion on the coated portion subtracted from the amount of corrosion on the uncoated portion. Does this random sample provide sufficient reason to conclude that the coating is beneficial? Use a = 0.01 and assume normality. n- 40, Σd- 207, Σα? = 6199 (a) Find t. (Give your answer correct to two decimal places.) (ii) Find the p-value. (Give your answer correct to four decimal places.) (b) State the appropriate conclusion. O Reject the null hypothesis, there is significant evidence that the coating is beneficial. Reject the null hypothesis, there is not significant evidence that the coating is beneficial. Fail to reject the null hypothesis, there is significant evidence that the coating is beneficial. Fail to reject the null hypothesis, there is not significant evidence that the coating is beneficial.Some frozen food dinners were randomly selected from this week's production and destroyed in order to measure their actual calorie content. The claimed calorie content is 200. Here are the calorie counts for each frozen dinner selected: 191 189 198 210 207 203 209 215 209 204 200 194 Assume the distribution of calories is normal. (a) The test statistic (z/t) is Use two decimals. (b) Does the sample indicate that the mean calorie content is 200? Set a = (c) Find the P-value of the test in (a)-(b). 0.07 ? Yes NoHello! I need answer for letter F only using Anderson-Darling test to support the normal distribution of the data.
- A sample of size 8 was collected from an unknown population 0 1126 10 11 20 1. Use the table of expected z-scores (rounded to the nearest tenth) below to construct the normality plot with expected z-scores on the x-axis and the observations on the y-axis: 5 -1.2 -0.5 0 0.5 1.2 Clear All Draw: 22- 2 not normal normal 19 # 6 -1.3 -0.6 -0.2 0.2 0.6 1.3 7 -1.4 -0.8 -0.4 0 0.4 0.8 1.4 8 -1.4 -0.9 -0.5 -0.2 0.2 0.5 0.9 1.4 9 -1.5 -0.9 -0.6 -0.3 0 0.3 0.6 0.9 1.5 2. Based on whether the pattern above is linear or not, in your opinion, was the sample drawn from a normal population or not?Data on the weights (lb) of the contents of cans of diet soda versus the contents of cans of the regular version of the soda is summarized to the right. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b) below. Use a 0.05 significance level for both parts. Diet Regular μ μ1 μ2 n 20 20 x 0.78646 lb 0.80233 lb s 0.00449 lb 0.00755 lb a. Test the claim that the contents of cans of diet soda have weights with a mean that is less than the mean for the regular soda. What are the null and alternative hypotheses? A. H0: μ1=μ2 H1: μ1≠μ2 B. H0: μ1=μ2 H1: μ1>μ2 C. H0: μ1≠μ2 H1: μ1<μ2 D. H0: μ1=μ2 H1: μ1<μ2 The test statistic, t, is nothing. (Round to two decimal places as needed.) The P-value is…The corrosive effects of various soils on coated and uncoated steel pipe was tested by using a dependent sampling plan. The data collected are summarized below, where d is the amount of corrosion on the coated portion subtracted from the amount of corrosion on the uncoated portion. Does this random sample provide sufficient reason to conclude that the coating is beneficial? Use ? = 0.01 and assume normality. n = 36, Σd = 227, Σd2 = 6244(a) Find t. (Give your answer correct to two decimal places.)(ii) Find the p-value. (Give your answer correct to four decimal places.)(b) State the appropriate conclusion. Reject the null hypothesis, there is significant evidence that the coating is beneficial.Reject the null hypothesis, there is not significant evidence that the coating is beneficial. Fail to reject the null hypothesis, there is significant evidence that the coating is beneficial.Fail to reject the null hypothesis, there is not significant evidence that the coating is beneficial.
- Answer the following: 1. Test Stat 2. Critical Region: Select one: a. T1.761 c. T>2.14 d. T>abs(2.145) 3. Conclusion: At 5% level of significance… Select one: a. the new type of fabric does not exceed the WRA of the top current brand b. the new type of fabric exceeds the WRA of the top current brandA study was carried out to compare mean customer satisfaction scores at service centers in city A, in city B, and in city C. The sample means on a scale of 0 to 10 were 8.4 in city A, 8.6 in city B, and 7.9 in city C. Each sample size = 100, MS error = 0.41, and the F test statistic = 27.4 has P-value <0.001. Complete parts a through d. a. What is the margin of error for separate 95% confidence intervals? (For df = 297, 1.025 = 1.968.) margin of error = *** tv (Round to two decimal places as needed.) Clear all Check answer ST AWomen are recommended to consume 1790 calories per day. You suspect that the average calorie intake is smaller for women at your college. The data for the 15 women who participated in the study is shown below: 1959, 1908, 1789, 1544, 1805, 1656, 1899, 1811, 1686, 1688, 1687, 1943, 1612, 1789, 1467 Assuming that the distribution is normal, what can be concluded at the α = 0.05 level of significance? For this study, we should use The null and alternative hypotheses would be: H0: H1: The test statistic = (please show your answer to 3 decimal places.) The p-value = (Please show your answer to 4 decimal places.) The p-value is α Based on this, we should the null hypothesis. Thus, the final conclusion is that ... The data suggest the population mean is not significantly less than 1790 at α = 0.05, so there is sufficient evidence to conclude that the population mean calorie intake for women at your college is equal to 1790. The data suggest the populaton…