Express the confidence interval 172.3 < µ < 229.1 in the form of ± ME. * + ME = 土 2.
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- A laboratory in Oregon is interested in finding the mean chloride level for a healthy resident in the state. A random sample of 60 mEq healthy residents has a mean chloride level of 103 If it is known that the chloride levels in healthy individuals residing in mEq find a 95% confidence interval for the true mean chloride level of all healthy Oregon Oregon have a standard deviation of 42 L residents. Then give its lower limit and upper limit. Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. (If necessary, consult a list of formulas.) Lower ?. limit: Upper limit:You are performing a two-tailed z-test If a = 0.2, find the positive critical value, to two decimal places.A simple random sample from a population with a normal distribution of 108 body temperatures has x= 98.50 F and s 0.62°F. Construct a 98% confidence interval estimate of the standard deviation of body temperature of all healthy humans. Click the icon to view the table of Chi-Square critical values. O'FTo obtain information on the corrosion-resistance properties of a certain type of steel conduit, 45 specimens are buried in soil for a 2-year period. The maximum penetration (in mils) for each specimen is then measured, yielding a sample average penetration of x = 53.9 and a sample standard deviation of s = 4.5. The conduits were manufactured with the specification that true average penetration be at most 50 mils. They will be used unless it can be demonstrated conclusively that the specification has not been met. What would you conclude? (Use a = 0.05.) State the appropriate null and alternative hypotheses. O Ho: H = 50 H3: µ # 50 Ο H: μ 50 H3:µ > 50 Ο H: μ= 50 Hz: µ > 50 Ο H: μ > 50 H: µ = 50 Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.) Z = P-value = State the conclusion in the problem context. O Reject the null hypothesis. There is sufficient evidence to conclude that the true…A recent study by Rover.com surveyed a random sample of 1500 dog owners. The research revealed that 975 of the dog owners admit to taking more photos of their dog than their significant other. Which of the following gives a 95% confidence interval for the proportion of all dog owners that would admit to taking more photos of their dog than their significant other? (A) 0.65 \pm1.96 \sqrt{\frac{0.65(0.35)}{1500}}0.65±1.9615000.65(0.35) (B) 0.65 \pm 2.576 \sqrt{\frac{0.65(0.35)}{1500}}0.65±2.57615000.65(0.35) (C) 0.65 \pm1.96 \frac{0.65}{\sqrt{1500}}0.65±1.9615000.65 (D) 975 \pm 2.576 \frac{975}{\sqrt{1500}}975±2.5761500975 (E) 975 \pm1.96 \sqrt{\frac{975(525)}{1500}}975±1.961500975(525)Express the confidence interval (0.071,0.133) in the form of ^p−E< p < ^p+E. < p <A random sample of 22 commercials made since the year 2010 have a mean length of 25 seconds with a standard deviation of 14 seconds. The margin of error that corresponds to a 90% confidence interval for the mean length of all commercials made since 2010 is A similar study to the one above was conducted, and the interval was found to be (14.1, 24.1). Can we claim that the mean length of all commercials made since 2010 is more than 15 seconds?The lifetime of a certain brand of electric light bulb is known to have a standard deviation of 53 hours. Suppose that a random sample of 50 bulbs of this brand has a mean of 495 hours. Find a 99% confidence interval for the true mean lifetime of all light bulbs of this brand. Then give the lower and upper limit. Carry intermediate computations to 3 decimal places, and answers to 1.1. In the past, a chemical company produced 880 pounds of a certain type of plastic per day. Now, using a newly developed and less expensive process, the mean daily yield of plastic for the first 50 days of production is 871 pounds; the standard deviation is 21 pounds. Do the data provide sufficient evidence to indicate that the mean daily yield for the new process is less than that of the old procedure? (Use α=0.05) (d) Conclusion of the test above is Reject the null hypothesis and the mean daily yield for the new process is less than that of the old procedure. Reject the null hypothesis and the mean daily yield for the new process is not less than that of the old procedure. Do not reject the null hypothesis and the mean daily yield for the new process is less than that of the old procedure. Do not reject the null hypothesis and the mean daily yield for the new process is not less than that of the old procedure.Data taken from a sample of 25 LED lamps, a sample average of 2 years was obtained and a standard deviation of s=0.4 years. The estimator of the mean of lifetime of LED lamps at the confidence level/coefficient (1-α) is (2 ± 0.05 talpha/2 , where talpha/2 is the value of t in the t distribution with v=24) (2 ± 0.10 talpha/2 , where talpha/2 is the value of t in the t distribution with v=24) (2 ± 0.08 talpha/2 , where talpha/2 is the value of t in the t distribution with v=24 (other)An machine with constant failure rate 0.002 per hours is operated 8 hours per day, 230 days per year. The mean downtime required to repair the machine and bring it back into operation is MDT = 5 hours. The machine can only fail during active operation. If a repair action cannot be completed within normal working hours, overtime will be used to complete the repair such that the machine is available next morning. (a) Determine the average availability of the machine (during the planned working hours). (b) Determine the average availability of the machine if the use of overtime were not allowed.A researcher studying the lifespan of a certain species of bacteria. A preliminary sample of 25 bacteria reveals a sample mean of = 70 hours with a standard deviation of s = 7 hours. He would like to estimate the mean lifespan for this species of bacteria to within a margin of error of 0.5 hours at a 99% level of confidence. What sample size should you gather to achieve a 0.5 hour margin of error? He would need to sample bacteria.
