Question 2
A manufacturer of sprinkler systems used for fire protection in office buildings claims that the true average system-activation temperature is 130 degrees. A sample of 9 systems when tested yields a sample average activation temperature of 131.08 degrees. If the distribution of activation times is normal with standard deviation 1.5 degrees, test at the 1% level of significance to see if the data shows evidence that is different from the manufacturers claim. (a) State the null and alternative hypotheses for the test. [2 marks] (b) What is the test statistic? [1 mark](c) Calculate the value of the test statistic for this test. [2 marks] (d) Calculate the p-value for this test. [2 marks] (e) State the conclusion of this test. Give a reason for your answer. [3 marks] Total 10 marks
Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 130
Alternative Hypothesis, Ha: μ ≠ 130
Test statistic,
z = (xbar - mu)/(sigma/sqrt(n))
z = (131.08 - 130)/(1.5/sqrt(9))
z = 2.16
P-value Approach
P-value = 0.0308
As P-value >= 0.01, fail to reject null hypothesis.
not different than population mean
Question 2 A manufacturer of sprinkler systems used for fire protection in office buildings claims that...
5. A manufacturer of sprinkler systems used for fire protection in office buildings claims that the true average system-activation temperature is 130°F. A sample of 9 systems, when tested, yields a sample average activation temperature of 131.08°F. Assume that the distribution of activation times is normal with a population standard deviation of 1.5°F. a) Does the data contradict the claim at 0.01 level? Carry out a detailed test by writing the appro- priate hypotheses, rejection region and conclusion. [4] b)...
5. A manufacturer of sprinkler systems used for fire protection in office buildings claims that the true average system-activation temperature is 130°F. A sample of 9 systems, when tested, yields a sample average activation temperature of 131.08°F. Assume that the distribution of activation times is normal with a population standard deviation of 1.5°F. a) Does the data contradict the claim at 0.01 level? Carry out a detailed test by writing the appro- priate hypotheses, rejection region and conclusion. [4] b)...
5. A manufacturer of sprinkler systems used for fire protection in office buildings claims that the true average system-activation temperature is 130°F. A sample of 9 systems, when tested, yields a sample average activation temperature of 131.08°F. Assume that the distribution of activation times is normal with a population standard deviation of 1.5°F. a) Does the data contradict the claim at 0.01 level? Carry out a detailed test by writing the appro- priate hypotheses, rejection region and conclusion. [4 b)...
5. A manufacturer of sprinkler systems used for fire protection in office buildings claims that the true average system-activation temperature is 130°F. A sample of 9 systems, when tested, yields a sample average activation temperature of 131.08°F. Assume that the distribution of activation times is normal with a population standard deviation of 1.5°F. a) Does the data contradict the claim at 0.01 level? Carry out a detailed test by writing the appro- priate hypotheses, rejection region and conclusion. [4] b)...
5. A manufacturer of sprinkler systems used for fire protection in office buildings claims that the true average system-activation temperature is 130°F. A sample of 9 systems, when tested, yields a sample average activation temperature of 131.08°F. Assume that the distribution of activation times is normal with a population standard deviation of 1.5°F. a) Does the data contradict the claim at 0.01 level? Carry out a detailed test by writing the appro- priate hypotheses, rejection region and conclusion. [4] b)...
A manufacturer of sprinkler systems used for fire protection in office buildings claims that the true average system-activation temperature is 130oF. A sample of n = 9 systems, when tested, yields a sample average activation temperature of 131.08oF. Suppose the distribution of activation temperature is normal with standard deviation 3.0 oF. Denote the true average system-activation temperature by µ ( oF). Consider testing H0 : µ = 130 versus Ha : µ 6= 130 at the significance level α =...
A manufacturer of sprinkler systems used for re protection in once buildings claims that the true 130F. average system-activation temperature is A sample of 9 systems when tested, yields a sample 131:08F. average activation temperature of Assume that the distribution of activation times is normal 1:5F. with a population standard deviation of a) Does the data contradict the claim at 0.01 level? Carry out a detailed test by writing the appropriate hypotheses, rejection region and conclusion. b) Find a 95%...
Question 2 4 points ✓ Saved A manufacturer of sprinkler systems designed for fire protection claims that the average activating temperature is 135 degrees. To test this claim, a (type in "left","right", or "two") -tailed hypothesis should be set up, A random sample of 32 systems has a mean activation temperature 134 degrees with a standard deviation of 3.3 degrees. What is the P-value for this sample? Round to the nearest thousandth. At significant level 0.1, Is there enough evidence...
a)A manufacturer of sprinkler systems designed for fire protection claims that the average activating temperature is 135 degrees. To test this claim, a (type in "left", "right", or "two") -tailed hypothesis should be set up. b)A random sample of 32 systems has a mean activation temperature 134 degrees with a standard deviation of 3.3 degrees. What is the P-value for this sample? Round to the nearest thousandth. c)At significant level 0.1, is there enough evidence to reject the manufacturer's clam?...
Question 8 2 pts A sprinkler manufacturer claims that the average activating temperatures is at least 134 degrees. To test this claim, you randomly select a sample of 32 systems and find the mean activation temperature to be 133 degrees. Assume the population standard deviation is 3.3 degrees. Find the standardized test statistic and the corresponding p-value O z-test statistic - -1.71, p-value-0.0432 O z-test statistic -1.71, p-value -0.0865 O z-test statistic -1.71, p-value- 0.0865 O z-test statistic - 1.71,...