For the hypothesis test Ho:μ = 5 against H₁:μ< 5 and variance known, calculate the P-value for the following test statistic: Zo = - 1.64. Round your answer to three decimal places (e.g. 98.765). P-value = 1
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- You wish to test the following claim (H,) at a significance level of a = 0.10. H.:u = 53.6 Ha:u + 53.6 You believe the population is normally distributed, but you do not know the standard deviation. You obtain a sample of size n = 105 with mean M = 55.6 and a standard deviation of SD = 14.4. What is the test statistic for this sample? (Report answer accurate to three decimal places.) test statistic = What is the p-value for this sample? {Report answer accurate to four decimal places.) P-value = The p-value is.. O less than (or equal to) o O greater than a This test statistic leads to a decision to... O reject the null O accept the null O fail to reject the null As such, the final conclusion is that.. O There is sufficient evidence to warrant rejection of the claim that the population mean is not equal to 53.6. O There is not sufficient evidence to warrant rejection of the claim that the population mean is not equal to 53.6. O The sample data support the claim that the population mean…You wish to test the following claim (HaHa) at a significance level of α=0.05 Ho:μ=59.9 Ha:μ>59.9You believe the population is normally distributed, but you do not know the standard deviation. You obtain a sample of size n=93n=93 with mean M=62.6M=62.6 and a standard deviation of SD=9.4SD=9.4.What is the test statistic for this sample? (Report answer accurate to three decimal places.)test statistic = What is the p-value for this sample? (Report answer accurate to four decimal places.)p-value = The p-value is... less than (or equal to) αα greater than αα This test statistic leads to a decision to... reject the null accept the null fail to reject the null As such, the final conclusion is that... There is sufficient evidence to warrant rejection of the claim that the population mean is greater than 59.9. There is not sufficient evidence to warrant rejection of the claim that the population mean is greater than 59.9. The sample data support the claim that the…Two accounting professors decided to test if the variance of their grades is different or no. To accomplish this, they each graded the same 10 exams with the following results: Standard Deviation 22.4 Mean Grade Professor 1 79.3 Professor 2 82.1 12.0 F What is the P-value of the test? Select one: a. between 0.01 and 0.025 Ob. between 0.025 and 0.05 C. between 0.05 and 0.10 Od. bigger than 0.10
- You wish to test the following claim (Ha) at a significance level of a = 0.001. H.:µ = 59.3 Ha:µ + 59.3 You believe the population is normally distributed, but you do not know the standard deviation. You obtain a sample of size n = 666 with mean M = 57.9 and a standard deviation of SD = 15.4. What is the test statistic for this sample? (Report answer accurate to three decimal places.) test statistic = What is the p-value for this sample? (Report answer accurate to four decimal places.) p-value = The p-value is.. less than (or equal to) a Ogreater than a This test statistic leads to a decision to... reject the null |accept the null fail to reject the null As such, the final conclusion is that... There is sufficient evidence to warrant rejection of the claim that the population mean is not equal to 59.3. There is not sufficient evidence to warrant rejection of the claim that the population mean is not equal to 59.3. The sample data support the claim that the population mean is not equal…You wish to test the following claim (HA) at a significance level of a = 0.002. H.:µ = 76.5 Ha: µ > 76.5 You believe the population is normally distributed, but you do not know the standard deviation. You obtain a sample of size n = 16 with mean M = 95.2 and a standard deviation of SD = 18.2. What is the test statistic for this sample? (Report answer accurate to three decimal places.) test statistic = 4.1099 V What is the p-value for this sample? (Report answer accurate to four decimal places.) p-value =A scrap metal dealer claims that the mean of his cash sales is 'no more than R80', but an Internal Revenue Service agent believes the dealer is untruthful. The agent randomly selects a sample of 20 cash customers and find the mean purchase to be R91, with a variance of R441. The test statistic (rounded off to four decimals) is: A. T = 2.3425 B. Z = 2.3425 C. T = 0.0113 D. Z = 0.2381
- The corrosive effects of various soils on coated and uncoated steel pipe was tested by using a dependent sampling plan. The data collected are summarized below, where d is the amount of corrosion on the coated portion subtracted from the amount of corrosion on the uncoated portion. Does this random sample provide sufficient reason to conclude that the coating is beneficial? Use ? = 0.01 and assume normality. n = 36, Σd = 227, Σd2 = 6244(a) Find t. (Give your answer correct to two decimal places.)(ii) Find the p-value. (Give your answer correct to four decimal places.)(b) State the appropriate conclusion. Reject the null hypothesis, there is significant evidence that the coating is beneficial.Reject the null hypothesis, there is not significant evidence that the coating is beneficial. Fail to reject the null hypothesis, there is significant evidence that the coating is beneficial.Fail to reject the null hypothesis, there is not significant evidence that the coating is beneficial.I need the answer as soon as possibleJustin wants to know whether a commonly prescribed drug does improve the attention span of students with attention deficit disorder (ADD). He knows that the mean attention span for students with ADD who are not taking the drug is 2.3 minutes long. His sample of 12 students taking the drug yielded a mean of 4.6 minutes. Justin can find no information regarding σx , so he calculated s2x =1.96. a. Identify the independent and dependent variables. b. In a sentence, state the null hypothesis being tested. c. Using symbols, state the null and alternative hypotheses. d. Determine the critical region using a one-tailed test with alpha = .05. e .Conduct the hypothesis test (Do the math and compare the t-critical and t-obtained values). f. State your conclusions in terms of H0 (Should you reject the H0 or fail to reject/accept the H0). g. Based on your analysis, is there a relationship between the drug and attention span?
- Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 4.6 parts/million (ppm). A researcher believes that the current ozone level is at an excess level. The mean of 14 samples is 4.9 ppm with a variance of 1.2 Does the data support the claim at the 0.01 level? Assume the population distribution is approximately normal. Step 2 of 5 : Find the value of the test statistic. Round your answer to three decimal places.During an economic crisis, the average value of homes in a community of 36 homes lost $9445 with a standard deviation of $1300. The average home value in the region lost $8990. Was this community of 36 homesunusual? Use a t-test to decide if the average loss observed was significantly different from the region value. Use a level of significance α=0.05. Identify the hypotheses for the test. H0: μ ▼ not equals≠ equals= greater than> less than< nothing HA: μ ▼ equals= greater than> not equals≠ less than< nothing The test statistic is nothing. (Round to two decimal places as needed.) The P-value is nothing. (Round to three decimal places as needed.) What is the conclusion of the test? Was the average loss observed significantly different from the region value? ▼ Reject Fail to reject the null hypothesis because the P-value is ▼ greater less than the given significance level, and conclude that the…During an economic crisis, the average value of homes in a community of 36 homes lost $9284 with a standard deviation of $1450. The average home value in the region lost $8790. Was this community of 36 homes unusual? Use a t-test to decide if the average loss observed was significantly different from the region value. Use a level of significance α=0.05. Identify the hypotheses for the test. H0: μ ▼ less than< greater than> equals= not equals≠ nothing HA: μ ▼ less than< not equals≠ greater than> equals= nothing The test statistic is nothing. (Round to two decimal places as needed.) The P-value is nothing. (Round to three decimal places as needed.) What is the conclusion of the test? Was the average loss observed significantly different from the region value? ▼ Fail to reject Reject the null hypothesis because the P-value is ▼ greater less than the given significance level, and conclude that the…