Based on a large sample of capacitors of a certain type, a 95% confidence interval for the mean capacitance, in μF, was computed to be (0.217, 0.257). Find a 90% confidence interval for the mean capacitance of this type of capacitor. (Round the final answers to three decimal places.)
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- A chemical manufacturer claims that its ice-melting product will melt snow and ice at temperatures as low as 0◦ Fahrenheit. A representative for a large chain of hardware stores is interested in testing this claim. The chain purchases a large shipment of 5-pound bags for distribution. The representative wants to know, with 95% confidence and within ±0.05, what proportion of bags of the chemical perform the job as claimed by the manufacturer. How many bags does the representative need to test? What assumption should be made concerning the population proportion? (This is called destructive testing; that is, the product being tested is destroyed by the test and is then unavailable to be sold.) The population proportion should be assumed to be equal to ________+_ . (Round to four decimal places as needed.)Heights (cm) and weights (kg) are measured for 100 randomly selected adult males, and range from heights of 133 to 188 cm and weights of 40 to 150 kg. Let the predictor variable x be the first variable given. The 100 paired measurements yield x=167.80 cm, y=81.36 kg, r=0.286, P-value=0.004, and y=−108+1.19x. Find the best predicted value of y (weight) given an adult male who is 138 cm tall. Use a 0.10 significance level.A standardized IQ test is known to be normally distributed and has μ=100 and σ=10. there is a common belief that students in wealthier school districts tend to have higher IQ scores. A researcher randomly selects a sample of 25 students from wealthy school districts. The average IQ of the sample is x̅=103. Conduct a test if the average score for students from wealthy districts is higher than 100, assuming α=0.05
- A nutritionist claims that the mean tuna consumption by a person is 3.4 pounds per year. A sample of 90 people shows that the mean tuna consumption by a person is 3.3 pounds per year. Assume the population standard deviation is 1.02 pounds. At α=0.03, can you reject the claim? (a) Identify the null hypothesis and alternative hypothesis. A. H0: μ=3.4 Ha: μ≠3.4 Your answer is correct. B. H0: μ>3.4 Ha: μ≤3.4 C. H0: μ≤3.3 Ha: μ>3.3 D. H0: μ≤3.4 Ha: μ>3.4 E. H0: μ>3.3 Ha: μ≤3.3 F. H0: μ≠3.3 Ha: μ=3.3 (b) Identify the standardized test statistic. z=negative 1.34−1.34 (Round to two decimal places as needed.)Use the information below to answer Questions #16 - 22. The article "Probabilistic Fatigue Evaluation of Riveted Railway Bridges" (J. of Bridge Engr., 2008: 237- 244) suggested the exponential distribution with A = 9 as a model for the distribution of stress range (MPa) in certain bridge connections. What proportion of stress ranges are at least 2 MPa? Round to four decimal places as needed.Heights (cm) and weights (kg) are measured for 100 randomly selected adult males, and range from heights of 138 to 194 cm and weights of 39 to 150 kg. Let the predictor variable x be the first variable given. The 100 paired measurements yield x=167.95 cm, y=81.57 kg, r=0.362, P-value=0.000, and y=−103+1.07x. Find the best predicted value of y (weight) given an adult male who is 180 cm tall. Use a 0.01 significance level. The best predicted value of y for an adult male who is 180 cm tall is= kg.
- A subject with an alpha=0.25 has an action-value for symbol A = 0.6 and symbol B = 0.75. The subject selects A and receives a reward of 1. a. Calculate the prediction error (delta) for this trial b. Calculate the new action-value for symbol A (use the Rescorla-Wagner equation) c. After this trial, which symbol is the subject more likely to select.12c: One researcher estimated that smoking reduces the sense of smell. The mean of the odor test for non-smokers is μ=18.4. The following scores were obtained by giving the same odor test to an unbiased sample of a cigarette smokers: 16 14 19 17 16 17 15 18 19 12 14 18 c) What is the critical t value for alpha=.05?For an alpha = .025 and degrees of freedom, df = 15, what is the one-tailed t-value? Give your answer exactly as it appears in the table, with three decimal places.
- Question 4 The Department of Energy wants to determine the percentage of mining firms that follow energy saving production processes. A random sample of 250 mining firms was selected and it was found that 95 out of 250 firms follow these practices. Estimate with 99% confidence, the percentage interval of mining firms that practice the energy saving production processes. Use a critical value of 2,58.You want to obtain a sample to estimate a population mean. Based on previous evidence, you believe the population standard deviation is approximately σ=77.5σ=77.5. You would like to be 99% confident that your esimate is within 0.5 of the true population mean. How large of a sample size is required?n = Do not round mid-calculation. However, use a critical value accurate to three decimal places — this is important for the system to be able to give hints for incorrect answers.A nutritionist claims that the mean tuna consumption by a person is 3.8 pounds per year. A sample of 70 people shows that the mean tuna consumption by a person is 3.5 pounds per year. Assume the population standard deviation is 1.04 pounds. At α=0.09, can you reject the claim? (a) Identify the null hypothesis and alternative hypothesis. A. H0: μ=3.8 Ha: μ≠3.8 B. H0: μ≠3.5 Ha: μ=3.5 C. H0: μ>3.5 Ha: μ≤3.5 D. H0: μ>3.8 Ha: μ≤3.8 E. H0: μ≤3.5 Ha: μ>3.5 F. H0: μ≤3.8 Ha:μ>3.8 b) Identify the standardized test statistic. z= (Round to two decimal places as needed.) (c) Find the P-value. (d) Decide whether to reject or fail to reject the null hypothesis. A. Reject H0. There is not sufficient evidence to reject the claim that mean tuna consumption is equal to 3.8pounds. B. Reject H0. There is sufficient evidence to reject the claim that mean tuna consumption is equal to…