(a) State the hypotheses and identify the claim. H₂: (Choose one) (Choose one) This hypothesis test is a (Choose one) test. 0<0 000-0 D-D X H Ś
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- Obtaining the P-value. Let x denote the test statistic for a hypothesis test and x0 its observed value. Then the P-value of the hypothesis test equals a. P(x ≥ x0) for a right-tailed test,b. P(x ≤ x0) for a left-tailed test,c. 2 min{P(x ≤ x0), P(x ≥ x0)} for a two-tailed test, where the probabilities are computed under the assumption that the null hypothesis is true. Suppose that you are considering a one-mean z-test. Verify that the probability expressions in parts (a)–(c) are equivalent to those obtained in Exercise. Consider a one-mean z-test. Denote z0 as the observed value of the test statistic z. If the test is right tailed, then the P value can be expressed as P(z ≥ z0). Determine the corresponding expression for the P-value if the test is a. left tailed. b. two tailed.State whether the standardized test statistict indicates that you should reject the null hypothesis. Explain. (a) t=2.242 (b) t= 0 (c) t=2.195 (d) t= -2.254 0. (a) For t 2.242, should you reject or fail to reject the null hypothesis? O A. Reject Ho, because t> 2.219. O B. Reject Ho, because t 2.219. O D. Fail to reject Ho, because t<2.219.#15 Thanks
- The percentage of viewers tuned to 60 minutes is equal to 69%. Is the original claim the null or alternative hypothesis? null alternative What are the correct hypotheses? (Select the correct symbols.)H0: Select an answer p̂ σ s σ² p μ s² x̄ ? < ≤ ≥ > ≠ = Select an answer 1% 0.01 50% 0.1 69 0.95 69% 0.69 0.05 10% 0.5 H1: Select an answer σ² x̄ s² μ s σ p p̂ ? ≠ < ≥ ≤ > = Select an answer 0.5 0.69 0.1 50% 69 10% 69% 1% 0.01 0.05 0.95 Submit QuestionQuestion 7At a magic shop, the salesperson shows you a coin that he says will land on heads more than 70% of the times it is flipped. In an attempt to convince you he's correct, the salesperson asks you to try the coin yourself. You flip the coin 65 times. (Consider this a random sample of coin flips.) The coin lands on heads 49 of those times. Complete the parts below to perform a hypothesis test to see if there is enough evidence, at the 0.10 level of significance, to support the salesperson's claim that the proportion, p, of all times the coin lands on heads is more than 70%. (a) State the null hypothesis H, and the alternative hypothesis H, that you would use for the test. Ho H: 0 D=D (b) For your hypothesis test, you will use a Z-test. Find the values of np andn (1-p) to confirm that a Z-test can be used. (One standard is that np 2 10 and n(1-p) > 10 under the assumption that the null hypothesis is true.) Here n is the sample size and p is the population proportion you are testing.The porportion of married women is greater than 53%. Is the original claim the null or alternative hypothesis? null alternative What are the correct hypotheses? (Select the correct symbols.)H0: Select an answer σ² μ x̄ σ p s² p̂ s ? ≤ > ≠ = ≥ < Select an answer 53 0.1 0.5 0.05 53% 0.53 50% 10% 0.95 0.01 1% H1: Select an answer x̄ p̂ p σ² μ σ s s² ? > ≤ = < ≠ ≥
- I need this urgently pleaseA large nationwide poll recently showed an unemployment rate of 9% in the US. The mayor of a local town wonders this national result holds true for her town, so she plans on taking a sample of her residents to see if the unemployment rate is significantly different than 9% in her town. Let T represent the unemployment rate in her town. Here are the hypotheses she'll use: Null Hypothesis: T = 0.09 %3D Alternative Hypothesis: T 0.09 Which type of Error has been committed in the following situation? • She concludes the town's unemployment rate is not significantly different to 9% when it is actually 9%. Type II Error Type I Error No ErrorBefore changes to its management staff, an automobile assembly line operation had a scheduled mean completion time of 14.2 minutes. The standard deviation of completion times was 1.6 minutes. An analyst at the company suspects that, under new management, the mean completion time, µ, is now less than 14.2 minutes. To test this claim, a random sample of 12 completion times under new management was taken by the analyst. The sample had a mean of 13 minutes. Assume that the population is normally distributed. Can we support, at the 0.01 level of significance, the claim that the population mean completion time under new management is less than 14.2 minutes? Assume that the population standard deviation of completion times has not changed under new management. Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places, and round your responses as specified below. (If necessary, consult a list of formulas.)
- q-10An engineer working for a large agribusiness has developed two types of soil additives he calls Add1 and Add2. The engineer wants to test whether there is any difference between the two additives in the mean yield of tomato plants grown using these additives. The engineer studies a random sample of 11 tomato plants grown using Add1 and random sample of 13 tomato plants grown using Add2. (These samples are chosen independently.) When the plants are harvested, he counts their yields. These data are shown in the table. sume Yields (in number of tomatoes) 97, 115, 126, 108, 111, 131, 90, 93, 136, 101, 99 k 7 ment Add1 Add2 154, 171, 89, 100, 168, 172, 101, 192, 133, 176, 141, 113, 198 Send data to calculator Send data to Excel tub 2 Assume that the two populations of yields are approximately normally distributed. Can the engineer conclude, at the 0.01 level of significance, that there is a difference between the population mean of the yields of tomato plants grown with Add1 and the…Completely abstract example: If the p-value is 0.0062 what is the conclusion (use a = 0.05)? Because the p-value is above the alpha level, we fail to reject the null hypothesis. Because the p-value is above the alpha level, we reject the null hypothesis. Because the p-value is below the alpha level, we fail to reject the null hypothesis. Because the p-value is below the alpha level, we reject the null hypothesis.