Acid rain, caused by the reaction of certain air pollutants with rainwater, appears to be a growing problem in Eastern Canada. (Acid rain affects the soil and causes corrosion on exposed metal surfaces.) Pure rain falling through clean air registers a pH value of 5.7 [pH is a measure of acidity: 0 is acid; 14 is alkaline). Suppose water samples from 50 rainfalls are analyzed for pH, and E and s are equal to 3.1 and 0.2, respectively. Find a 99% confidence interval for the mean pH in rainfall. (Round your answers to three decimal places.) 3024 x to 3176 x Interpret the interval. O 99% of all values will fall within the interval. 0 There is a 1% chance that an individual sample mean will tall within the interval. 0 There is a 99% chance that an individual sample mean will fall within the interval. In repeated sampling, 99% ofall intervals constructed in this manner will enclose the population mean. 0 In repeated sampling, 1% of all intervals constructed in this manner will enclose the population mean. V What assumption must be made for the confidence interval to be valid'.> The sample must be random. 0 The sampling distribution must be symmetrical. O The standard deviation must be less than 10. 0 There must be at least 100 samples. 0 The sample mean must be greater than S. A sample survey is designed to estimate the proportion of sports utility vehicles being driven in PEI. A random sample of 500 registrations are selected from a PEI database, and 84 are classified as sports utility vehicles. (a) Use a 95% confidence interval to estimate the proportion of sports utility vehicles in PEI. (Round your answers to three decimal places.) :i x tox (b) How can you estimate the proportion of sports utility vehicles in PEI with a higher degree of accuracy? (HINT: There are two answers. Select all that apply.) [3 increase the sample size n D decrease the sample size n D conduct a nonirandom sample increase Za/Z by increasing the confidence coefcient D decrease 2W2 by decreasing the confidence coefficient X Independent random samples were selected from populations 1 and 2. The sample sizes, means, and variances are as follows. Population 1 2 Sample Size 30 64 Sample Mean 12.3 8.2 Sample Variance 1.32 4.11 (a) Find a 95% confidence interval for estimating the difference in the population means (M, - /2). ( Round your answers to two decimal places.) 2.57 X to 5.63 X (b) Based on the confidence interval in part (a), can you conclude that there is a difference in the means for the two populations? Explain. Since the value My - M, = 0 is in the confidence interval, it is likely that there is a difference in the population means. Since the value , - M, = 0 is not in the confidence interval, it is not likely that there is a difference in the population means. O Since the value My - M, = 0 is in the confidence interval, it is not likely that there is a difference in the population means. O Since the value My - M, = 0 is not in the confidence interval, it is likely that there is a difference in the population means.A small amount of the trace element selenium, 507200 micrograms (pig) per day, is considered essential to good health. Suppose that random samples of HI = \"2 = .20 adults were selected from two regions of Canada and that a day's intake of selenium, from both liquids and solids, was recorded for each person. The mean and standard deviation of the selenium daily intakes for the 30 adults from region 1 were x1 : 167.1 and s1 : 24.9 pig, respectively. The corresponding statistics for the 30 adults from region 2 were ;2 : 140.1anc selenium intakes for the two regions. (Round your answers to three decimal places) is to S is Interpret this interval. 95% of all differences Will fall within the interval. 0 In repeated sampling, 5% of all intervals constructed in this manner will enclose the difference in population means. 0 There is a 5% chance that the difference between indiVidual sample means will fall within the interval. 0 In repeated sampling, 95% of all intervals constructed in this manner will enclose the difference in population means 0 There is a 95% chance that the difference between indiViduai sample means will fall within the interval. 52 : 17.8 pg. Find a 95% confidence interval for the difference (p1 71:2) in the mean