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The following observations were made on fracture toughness of base plate of 18% nickel maraging steel in
ksi in . Answer the problems assuming that the test results from this study is
1) Explain why the t-curve is more appropriate than the z-curve in this case.
◼ 2.2. Based upon this study, find the 95% confidence interval of the mean fracture toughness.
◼ 2.3. Find the 99% lower confidence bound.
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- 5Your company sources motors from two suppliers: supplier A and supplier B. Your application requiresvery high maximum torque. You wish to know if there is a significant difference in maximum torque forthe motors from supplier A vs. supplier B. You sample 10 motors from each supplier and measure themaximum torque. The sample means and sample standard deviations are shown the following table. Is there evidence to conclude that one supplier is better than the other at a 95% confidence level?The numbers of successes and the sample sizes for independent simple random samples from two populations are provided for a right-tailed test. Use a 90% confidence interval. Complete parts (a) through (d). X1 = 23, n, = 90, x2 = 18, n2 = 100, a = 0.05 Click here to view Click here to view table of areas under the standard normal curve for negative values of z. table of areas under the standard normal curve for positive values of z. a. Determine the sample proportions. Determine the sample proportion p P, = (Round to three decimal places as needed.)
- 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The ages of registered voters in Smith County are normally distributed with a population standard deviation of 3 years and an unknown population mean. A random sample of 18 voters is taken and results in a sample mean of 55 years. Find the margin of error for a 95% confidence interval for the population mean. z0.10z0.10 z0.05z0.05 z0.025z0.025 z0.01z0.01 z0.005z0.005 1.282 1.645 1.960 2.326 2.576 You may use a calculator or the common z values above. Round the final answer to two decimal places.A sample of 12 concrete specimens from supplier A were tested for their compressive strength. The results gave an average of 85 MPa and a standard deviation of 4 MPa. From supplier B, 10 specimens were similarly tested and gave an average compressive strength of 81 MPa and a standard deviation of 5 MPa. (a) Using procedures of Hypothesis Testing, can we conclude at the 0.05 level of significance that the compressive strength of concrete from A exceeds that from B by more than 2 MPa? Assume the populations to be approximately normal with equal variances. (b) Is the assumption of equal variances in (a) justifiable at an a = 0.10?
- Can you solve this problem about the confidence interval?A simple random sample of 30 men from a normally distributed population results in a standard deviation of 8.7 beats per minute. The normal range of pulse rates of adults is typically given as 60 to 100 beats per minute. If the range rule of thumb is applied to that normal range , the result is a standard deviation of 10 beats per minute. Use the sample results with a 0.05 significance level to test the claim that pulse rates of men have a standard deviation equal to 10 beats per minute. Complete parts a through d below . b. Compute the test statistic. x^2 =____ ( round to three decimal places as needed ) c. Find the p value ___ ( round to four decimal places as needed ) . d. State the conclusionA simple random sample of 30 men from a normally distributed population results in a standard deviation of 10.4 beats per minute. The normal range of pulse rates of adults is typically given as 60 to 100 beats per minute. If the range rule of thumb is applied to that normal range, the result is a standard deviation of 10 beats per minute. Use the sample results with a 0.05 significance level to test the claim that pulse rates of men have a standard deviation equal to 10 beats per minute. Complete parts (a) through (d) below. a. Identify the null and alternative hypotheses. Choose the correct answer below. O A. Ho: o = 10 beats per minute O B. Ho: o2 10 beats per minute H: o<10 beats per minute H,: o< 10 beats per minute OC. Ho: o 10 beats per minute O D. Ho: o = 10 beats per minute H4: 0 = 10 beats per minute H: o# 10 beats per minute b. Compute the test statistic. (Round to three decimal places as needed.) c. Find the P-value. P-value = (Round to four decimal places needed. d. State…
- "The mean value of the design load on a foundation element is 6,000 kN, and its standard deviation is 1,200 kN. The estimated mean value of the bearing capacity is 12,000 kN, and its standard deviation 3,000 kN. If both the load and the resistance are Normally distributed, compute (a) the value of the reliability index B and (b) the probability of failure.", if the load and the resistance are positively correlated with p = 0.5.Multi-sensor data loggers were attached to free-ranging American alligators in a study conducted by Y. Watanabe for the article “Behavior of American Alligators Monitored by Multi-Sensor Data Loggers”. The margin of error for a sample of 68 dives was 31.2 seconds. Assume the population standard deviation is 100 seconds. If for the next study, a confidence interval for μ is to have a margin of error of 25 seconds and a confidence level of 99%, determine the required sample size. (Round up to the nearest whole number.)The adhesive strength of medical monitoring device has a known standard deviation of 0.55 dyne cm. Tests on the first five prototypes of a new model yielded the following values of adhesive strength: 2.96, 5.67, 2.76, 1.26 and 4.21 dyne-cm'. a. What is the 95% confidence interval for the mean adhesion? b. If the manufacturer wants the confidence interval to be no wider than 0.62 dyne-cm2, how many 1. observations should they take?