Suppose we want to test the hypothesis of Ho : u 2 1 versus H. : u < 1, and a random sample of n = 49 gives mean i = -1 and standard deviation s = 6.4516. What is the %3D p-value of the test? (a) p-value=0.005 (b) p-value=0.010 (c) p-value=0.015 (d) p-value=0.020 (e) p-value=0.025
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- The calibration of a scale is to be checked by weighing a 10-kg test specimen 25 times. Suppose that the results of different weighings are independent of one another and that the weight on each trial is normally distributed with = 0.2 kg. Let μ denote the true average weight reading on the scale. (a) What hypotheses should be tested? (b) What is the p-value when = 9.85? What would you conclude at significance level a = 0.01?#1) Flying time on an old route between 2 cities has a mean value of µ =5.25 hours with standard deviation is o =0.6 hours. There are 36 flights on a new route where the mean is x = 4.90 hours. Does this indicate that the average flying time for the new route is less than 5.25 hours? Use a 5% level of significance.1) Your instructor tells you that past average scores for the final exam are 75%. But this semester he has been using online quizzes to help your class better prepare for the final. Your class scored the following on the final exam: mean= 80% with a 10% standard deviation. There are 49 people in your class. Using a one-sample t test, what is your t score for this class? Assuming α = .05, did the online quizzes help your class perform better?
- At the alpha = 0.01 level , what is the correct conclusion for this test? The daily temperatures in fall and winter months in Virginia have a mean of 62F. A meteorologist in southwest Virginia believes the mean temperature is colder in this area. The meteorologist takes a random sample of 30 daily temperatures from the fall and winter months over the last five years in southwest Virginia. The mean temperature for the sample is 59 degrees * F with a standard deviation of 6.21 degrees * F The meteorologist conducts a one -sample t-test for and calculates a P value of 0.007. The meteorologist should reject the null hypothesis since 0.007 < 0.01 . There is convincing evidence that the mean temperature in fall and winter months in southwest Virginia is less than 62 F. The meteorologist should reject the null hypothesis since 0.007 < 0.01 . There is not convincing evidence that the mean temperature in fall and winter months in southwest Virginia is less than 62 F. The meteorologist…A doctor believes women have been getting taller. She knows that 10 years ago the average height of women in her city was 63 inches. The standard deviation is unknown. She randomly samples 8 women from the town and finds their average height to be 64.5 inches, the standard deviation was 6.5. state the hypotheses(H0 and H1) What is your conclusion based on α=0.052tailSuppose the mean IQ score of people in a certain country is 100. Suppose the director of a college obtains a simple random sample of 39 students from that country and finds the mean IQ is 104.5 with a standard deviation of 13.6. Complete parts (a) through (d) below. (a) Consider the hypotheses Ho: µ= 100 versus H,: µ> 100. Explain what the director is testing. Perform the test at the a = 0.10 level of significance. Write a conclusion for the test. Explain what the director is testing. Choose the correct answer below. O A. The director is testing if the sample provided sufficient evidence that the population mean IQ score is actually not equal to 100. OB. The director is testing if the sample provided sufficient evidence that the population mean IQ score is actually greater than 100. O C. The director is testing if the sample provided sufficient evidence that the population mean IQ score is actually not greater than 100. O D. The director is testing if the sample provided sufficient…
- A sample of n=52 heights (in cm). of women yielded a mean of x=163.769 and a standard deviation of s=6.4319, variance v=41.3693 for the hypotheses H0:μ=163.769 H0:μ<163.769 CALCULATE THE P-VALUE (Perform a normality test on the data, Proposal of hypotheses of independent means,Perform a test for equality of variances)A sample of size 31 was gathered from high school seniors to estimate how many intended to attend the state university. It is known that the proportion of the population answering "yes" was 0.7. What are the mean and standard deviation, and standard error of the mean of this sample? mean = E(p) = %3D standard error = SD(p) (accurate to 3 decimal places)Suppose the mean IQ score of people in a certain country is 103. Suppose the director of a college obtains a simple random sample of 37 students from that country and finds the mean IQ is 107.4 with a standard deviation of 13.8. Complete parts (a) through (d) below. (a) Consider the hypotheses Ho: H= 103 versus H: u> 103. Explain what the director is testing. Perform the test at the a = 0.01 level of significance. Write a conclusion for the test. Explain what the director is testing. Choose the correct answer below. A. The director is testing if the sample provided sufficient evidence that the population mean IQ score is actually greater than 103. B. The director is testing if the sample provided sufficient evidence that the population mean IQ score is actually not equal to 103. O C. The director is testing if the sample provided sufficient evidence that the population mean IQ score is actually equal to 103. O D. The director is testing if the sample provided sufficient evidence 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)) = .639A sample of 55 measurements resulted in a sample mean, ?⎯⎯⎯=9.3x¯=9.3 and sample standard deviation ?=1.47s=1.47. Using ?=0.05α=0.05, test the null hypothesis that the mean of the population is 10.510.5 or less against the alternative hypothesis that the mean of the population, ?>10.5μ>10.5 by giving the following:(a) the degree of freedom (b) the test statistic The final conclusion is A. There is not sufficient evidence to reject the null hypothesis that ?=10.5μ=10.5. B. We can reject the null hypothesis that ?=10.5μ=10.5