Determine the critical values for these tests of a population standard deviation.
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- You wish to test Ho: μ = 87.5 versus Ha: 87.5 at a significance level of 0.10. You obtain a sample of size 21 with a mean of 84.5 and a standard deviation of 10.4. You believe that the population is normally distributed. Round your answers to three decimal places, and round any interim calculations to four decimal places. What is the test statistic? You wish to test Ho:μ = 63.9 versus Ha: μ< 63.9 at a significance level of 0.10. You obtain a sample of size 28 with a mean of 55.1 and a standard deviation of 16.7. You believe that the population is normally distributed. Round your answers to three decimal places, and round any interim calculations to four decimal places. What is the test statistic?Suppose you are interested in measuring the amount of time on average it takes you to makeyour commute to school. You have estimated that the average time is 38.4 minutes with a standard deviation of 5.362 minutes. Over 12 random days, you measure your commute time. Assuming that your estimated parameters are correct and is normally distributed. Answer the following questions: (a) What is the distribution of the sample mean X¯?(b) What is the mean and standard deviation of the sample mean X¯?(c) What is the probability that the sample mean for sample size 12 will be greater than 36.19 minutes?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…
- Do a two-sample test for equality of means assuming unequal variances. Calculate the p-value using Excel. (a-1) Comparison of GPA for randomly chosen college juniors and seniors: x⎯⎯1x¯1 = 4.5, s1 = .20, n1 = 15, x⎯⎯2x¯2 = 4.9, s2 = .30, n2 = 15, α = .025, left-tailed test.(Negative values should be indicated by a minus sign. Round down your d.f. answer to the nearest whole number and other answers to 4 decimal places. Do not use "quick" rules for degrees of freedom.) d.f. Not attempted t-calculated Not attempted p-value Not attempted t-critical Not attempted (a-2) Based on the above data choose the correct decision. Do not reject the null hypothesis Reject the null hypothesisrandom sample of 10 subjects have weights with a standard deviation of 10.9919 kg. What is the variance of their weights? Be sure to include the appropriate units with the result.Scores on a certain "IQ" test for 18-25 year olds are normally distributed. A researcher believes that the average IQ score for students at a certain NJ college is less than 110 points, and so wants to test this hypothesis. The researcher obtain a SRS of 45 student IQ scores from school records and found the mean of the 45 results was 108 with a sample standard deviation of 21. The level of significance (alpha) used for this problem is 0.05. What is the appropriate test statistic (Student must complete by showing by formula using the ap- propriate values in that formula "showing work" and the final answer and appropriate label)? O T test score = (108-110)/(21/sqrt(45)) = -.639 T test score (108-110)/(21/sqrt(45)) = .639 %3D OT test score = (110-108)/(21/sqrt(45)) = .639 %3D T test score (108-110)/(45/sqrt(21)) =-.2037 %3D 素
- In a left-tailed hypothesis for a population mean where the population standard deviation is unknown, the test statistic for a random sample size 18 was calculated to be -4.0545. Determine the P-value for the test. XThe mean test score for a simple random sample of n=100 students was =80. The population standard deviation of test scores is σ=15. Construct a 95% confidence interval for the population mean test score μ.Results of a research claim to predict wind speed at a place with a variance of 4 km/hr or less. A random sample of 24 measurements provides a sample variance of s2 = 4.9. Assuming that the population distribution of wind speed is approximately normal, check the validity of the claim at 5% level of significance using chi-square test.
- The average salary for American college graduates is $48,600. You suspect that the average is less for graduates from your college. The 49 randomly selected graduates from your college had an average salary of $46,486 and a standard deviation of $9,200. What can be concluded at the αα = 0.01 level of significance? For this study, we should use Select an answer t-test for a population mean z-test for a population proportion The null and alternative hypotheses would be: H0:H0: ? p μ Select an answer = ≠ > < H1:H1: ? μ p Select an answer > ≠ = < The test statistic ? z t = (please show your answer to 3 decimal places.) The p-value = (Please show your answer to 4 decimal places.) The p-value is ? > ≤ αα Based on this, we should Select an answer reject accept fail to reject the null hypothesis. Thus, the final conclusion is that .Scores on a certain "IQ" test for 18-25 year olds are normally distributed. A researcher believes that the average IQ score for students at a certain NJ college is less than 110 points, and so wants to test this hypothesis. The researcher obtain a SRS of 45 student IQ scores from school records and found the mean of the 45 results was 108 with a sample standard deviation of 21. The level of significance (alpha) used for this problem is 0.05/ What is the appropriate test statistic (Student must complete by showing by formula using the ap- propriate values in that formula "showing work" and the final answer and appropriate label)? O T test score = (108-110)/(45/sqrt(21)) = -.2037 O I test score = (108-110)/(21/sqrt(45)) = -.639 T test score = (108-110)/(21/sqrt(45)) = .639 I test score = (110-108)/(21/sqrt(45)) = .639The average salary for American college graduates is $45,100. You suspect that the average is less for graduates from your college. The 50 randomly selected graduates from your college had an average salary of $42,788 and a standard deviation of $5,940. What can be concluded at the αα = 0.05 level of significance? For this study, we should use Select an answer t-test for a population mean z-test for a population proportion The null and alternative hypotheses would be: H0:H0: ? μ p Select an answer > < = ≠ H1:H1: ? p μ Select an answer ≠ > = < The test statistic ? z t = (please show your answer to 3 decimal places.) The p-value = (Please show your answer to 4 decimal places.) The p-value is ? > ≤ αα Based on this, we should Select an answer fail to reject reject accept the null hypothesis. Thus, the final conclusion is that ... The data suggest that the population mean is not significantly less than 45,100 at αα = 0.05, so there is…