500g=0.5kg
=type two error
std error =std deviation/(n)1/2 =0.5/(50)1/2=0.0707
for =14.9
=P(X>14.9)=P(Z>(14.9-14.9)/0.0707)=P(Z>0.0)=0.5
A fisherman claims that the mean breaking strength of his fishing line is 15 kg with...
Please write the full derivation of the answers. Thank you! Q2) 1 Point for part a, 1 point for part b; total 2 points A company has developed a new fishing line, which the company claims has a mean breaking strength of 15 kilograms with a standard deviation of 0.5 kilogram. To test the hypothesis that u = 15 kilograms against the alternative that u < 15 kilograms, a random sample of 50 lines will be tested. The critical region...
A new type of rope has a mean breaking strength of 15 kilograms with a standard deviation of 0.5 kilogram. To test the hypothesis, a random sample of 50 lines are tested and the average breaking strength is found to be 14.8 kilograms. (a) Test the hypothesis, at the 0.01 level of significane the the mean is that u = 15 kilograms against the alternative that u < 15 kilograms. (b) Evaluate the P value of the test.
A new type of rope has a mean breaking strength of 15 kilograms with a standard deviation of 0.5 kilogram. To test the hypothesis, a random sample of 50 lines are tested and the average breaking strength is found to be 14.8 kilograms. (a) Test the hypothesis, at the 0.01 level of significane the the mean is that μ = 15 kilograms against the alternative that μ < 15 kilograms. (b) Evaluate the P value of the test.
QUESTION 4 A new type of rope has a mean breaking strength of 25 kilograms with a standard deviation of 3 kilogram. To test the hypothesis, a random sample of 50 lines are tested and the average breaking strength is found to be 22 kilograms. (a) Test the hypothesis, at the 0.05 level of significane the the mean is that u = 25 kilograms against the alternative that u < 25 kilograms. (b) Evaluate the P value of the test.
QUESTION A new type of rope has a mean breaking strength of 10 kilograms with a standard deviation of 1 kilogram. To test the hypothesis, a random sample of 20 lines are tested and the average breaking strength is found to be 8.5 kilograms. (a) Test the hypothesis, at the 0.01 level of significane the the mean is that - 10 kilograms against the alternative that p < 10 kilograms. (b) Evaluate the P value of the test. т т...
Sozcüklero volp SORU 4 A new type of rope has a mean breaking strength of 10 kilograms with a standard deviation of 1 kilogram. To test the hypothesis, a random sample of 20 lines are tested and the average breaking strength is found to be 8.5 kilograms. (a) Test the hypothesis, at the 0.01 level of significane the the mean is that = 10 kilograms against the alternative that p < 10 kilograms. (b) Evaluate the value of the test....
(16 points) Suppose the breaking strength of plastic bags is a Gaussian random variable Bags from company i have a mean strength of 8 kilograms and a variance of 1 kg2; Bags from company 2 have a mean strength of 9 kilograms and a variance of 0.5 kg' Assume we check the sample mean X1o of the breaking strength of 10 bags, and use X1o to determine whether a batch of bags comes from company 1 (null hypothesis Ho) or...
Question 7 The mean breaking strength of yarn used in manufacturing drapery material is required to be more than 100 psi. Past experience has indicated that the standard deviation of breaking strength is 2.9 psi. A random sample of 9 specimens is tested, and the average breaking strength is found to be 100.6 psi. Statistical Tables and Charts (a) Calculate the P-value. Round your answer to 3 decimal places (e.g. 98.765). If a = 0.05, should the fiber be judged...
A new treatment has been developed for a certain type of cement, resulting in a mean pressure resistance of 5000kg/cm² and a standard deviation of 130kg/cm². To verify the null hypothesis of p = 5000 against the alternative of u <5000, a random sample of 50 pieces of cement are examined. Determine the probability of committing a Type II error if = 4960 and a = 0.05. A. 0.298 B. 0.954 C. 0.082 D. 0.105 E. none of the preceding
A manufacturer claims his light bulbs have a mean life of 1800 hours. A consumer group wants to test if their light bulbs do not last as long as the manufacturer claims. They tested a random sample of 270 bulbs and found them to have a sample mean life of 1790 hours and a sample standard deviation of 60 hours. Assess the manufacturer's claim. a) What is the null hypothesis? Correct: y = 1800 Incorrect x = 1800 Incorrect x...