The breaking strengths of cables produced by a certain company are approximately normally distributed. The company announced that the mean breaking strength is 2130 pounds, with a variance of 33,672.25. A consumer protection agency claims that the actual variance is higher. Suppose that the consumer agency wants to carry out a hypothesis test to see if its claim can be supported. State the null hypothesis H, and the alternative hypothesis H, they would use for this test. H, :0 Oso D>o 020 ?
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Q: Ho: 0 H;: []
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Q: The breaking strengths of cables produced by a certain company are approximately normally…
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- Two researchers conduct separate studies to test Ho: p=0.50 against Ha: p0.50, each with n = 400. Researcher A gets 225 observations in the category of interest, and p=225/400 =0.5625. Researcher B gets 227 in the category of interest, and p=227/400 = 0.5675. Complete parts (a) through (e) below. A ... OB. Hypothesis tests that produce similar P-values always have the same conclusion. OC. Hypothesis tests that result in the same conclusion do not necessarily have the same P-value. D. A difference in conclusions between two hypothesis tests does not necessarily mean that the test with the result that is deemed "statistically significant" is actually much more significant than the test with the result that is deemed "not statistically significant." d. From part a, part b, and part c, explain why important information is lost by reporting the result of a test as "P-value ≤0.05" versus "P-value > 0.05," or as "reject Ho" versus "Do not reject Ho," instead of reporting the actual P-value.…Perform the indicated hypothesis test. Assume that the two samples are independent simple random samples selected from normally distributed populations. Also assume that the population standard deviations are equal (o1 = 02), so that %3D the standard error of the difference between means is obtained by pooling the sample variances. 19. A researcher was interested in comparing the resting pulse rates of people who exercise regularly and the pulse rates of those who do not exercise regularly. Independent simple random samples of 16 people who do not exercise regularly and 12 people who exercise regularly were selected, and the resting pulse rates (in beats per minute) were recorded. The summary statistics are as follows. Do Not Exercise| Do Exercise X1 = 73.3 beats/min x2 68.2 beats/min S1 = 10.7 beats/min s2 = 8.2 beats/min %3D %3D n2 = 12 n1 = 16 %3D Use a 0.025 significance level to test the claim that the mean resting pulse rate of people who do not exercise regularly is greater than…You may need to use the appropriate technology to answer this question. The following table contains observed frequencies for a sample of 200. RowVariable Column Variable A B C P 22 46 52 Q 28 24 28 Test for independence of the row and column variables using ? = 0.05. State the null and alternative hypotheses. H0: Variable P is not independent of variable Q.Ha: Variable P is independent of variable Q.H0: The column variable is independent of the row variable.Ha: The column variable is not independent of the row variable. H0: The column variable is not independent of the row variable.Ha: The column variable is independent of the row variable.H0: Variable P is independent of variable Q.Ha: Variable P is not independent of variable Q. Find the value of the test statistic. (Round your answer to three decimal places.) Find the p-value. (Round your answer to four decimal places.) p-value = State your conclusion. Reject H0. We conclude that there is an association…
- The breaking strengths of cables produced by a certain company are approximately normally distributed. The company announced that the mean breaking strength is 2140 pounds with a variance of 33,672.25. A consumer protection agency claims that the actual variance is higher. Suppose that the consumer agency wants to carry out a hypothesis test to see if its claim can be supported. State the null hypothesis Ho and the alternative hypothesis H, they would use for this test. Ho: 0 H: 0 Họ: OOne sample has M = 18 and a second sample has M = 14. If the pooled variance for the two samples is 16, what is the value of Cohen’s d?Cindy is a server at a local restaurant who makes, on average, $58 in tips per daytime shift. Cindy thinks she will make more if she moves to the nighttime shift. Her manager will choose five random nighttime shifts for Cindy to work. Cindy plans to conduct a significance test on the hypotheses H0: mu = 58; Ha: mu > 58 at the alpha = 0.05 level, where is the true mean amount of tips Cindy will make working nighttime shifts. Are the conditions for performing the test met? No, the 10% condition is not met. No, the sample size is too small, and a Normal distribution cannot be assumed. Yes, all conditions are met for performing a significance test about the mean. Yes, only the random condition must be met in order to perform a significance test. No, there is no way the manager could choose a random sample of shifts to work.Test the given hypothesis. Assume that the population is normally distributed and that the sample has been randomly selected. A manufacturer uses a new production method to produce steel rods. A random sample of 37 steel rods resulted in lengths with a mean a+1 and standard deviation of 4.7 cm. At the 0.10 significance level, test the claim that the new production method has mean 5.5 cm, which was the mean for the old method. a=2The records of a casualty insurance company show that, in the past, its clients have had a mean of 1.8 auto accidents per day with a variance of 0.0016. The actuaries of the company claim that the variance of the number of accidents per day is no longer equal to 0.0016. Suppose that we want to carry out a hypothesis test to see if there is support for the actuaries' claim. State the null hypothesis H, and the alternative hypothesis H, that we would use for this test. Ho: o = H: 0.0016 OUsing a long rod that has length m, you are goingto lay out a square plot in which the length of eachside is m. Thus the area of the plot will be m2.However, you do not know the value of m, soyou decide to make n independent measurementsX1, X2,... Xn of the length. Assume that each Xihas mean m (unbiased measurements) and variance s2.a. Show that X2 is not an unbiased estimatorfor m2. [Hint: For any rv Y, E(Y2) ¼V(Y) + [E(Y)]2. Apply this with Y ¼ X.]b. For what value of k is the estimator X2 - kS2unbiased for m2? [Hint: Compute E(X2 - kS2).]The records of a casualty insurance company show that, in the past, its clients have had a mean of 1.9 auto accidents per day with a variance of 0.0025 . The actuaries of the company claim that the variance of the number of accidents per day is no longer equal to 0.0025 . Suppose that we want to carry out a hypothesis test to see if there is support for the actuaries' claim. State the null hypothesis H0 and the alternative hypothesis H1 that we would use for this test. H0:H1:A nutritionist wants to investigate whether her new diet will be effective in helping women aged 30-40 to lose weight. She will use a paired sample to determine whether the mean weight of women before going on this diet is greater than the mean weight of women after being on this diet for two months. Determine the null and alternative hypotheses for the proposed hypothesis test. O A. Let μ₁ denote the mean weight of women before going on this diet and let μ₂ denote the mean weight of women who have been on this diet for two months. The null and alternative hypotheses are Ho: H₁ H2 and Ha: H₁ H₂. O D. Let μ₁ denote the mean weight of women before going on this diet and let μ₂ denote the mean weight of women who have been on this diet for two months. The null and alternative hypotheses are Ho: H₁ H2 and Ha: H₁ H₂.Do question 7,8 and 9 only The new study also carried out a test to determine whether the population proportion of unvaccinated school children contracting winter flu was higher than the population proportion of vaccinated school children. The Z test statistic to test this belief is found to be 1.874. The corresponding p-value is 0.0305 0.1212 0.3036 0.7724 Question 7 Suppose that the new study uses a level of significance of 0.05 to test the claim in Question 6. The probability of Type I error is 0.025 0.05 0.95 0.975 Question 8 Based on previous studies of school children who were vaccinated and contracted the flu, the time in hours that the flu symptoms last is assumed to follow a normal distribution with a mean of 20.7 hours and a standard deviation of 7.3 hours. The probability that a randomly selected school child has flu symptoms for more than 24 hours is 0.1628 0.3256 0.6744 0.8372 Question 9 Suppose that a random sample of 5 vaccinated school children is taken. Assuming the…SEE MORE QUESTIONSRecommended textbooks for youMATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons Inc
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