A food safety guideline is that the mercury in fish should be below 1 part per million (ppm). Listed below are the amounts of mercury (ppm) found in tuna sushi sampled at different stores in a major city. Construct a 90% confidence interval estimate of the mean amount of mercury in the population. Does it appear that there is too much mercury in tuna sushi? 0.51 0.82 0.10 0.89 1.35 0.52 0.82 What is the confidence interval estimate of the population mean p? ppm < u < ppm (Round to three decimal places as needed.) Does it appear that there is too much mercury in tuna sushi? O A. Yes, because it is possible that the mean is greater than 1 ppm. Also, at least one of the sample values exceeds 1 ppm, so at least some of the fish have too much mercury. O B. No, because it is not possible that the mean is greater than 1 ppm. Also, at least one of the sample values is less than 1 ppm, so at least some of the fish are safe. O C. No, because it is possible that the mean is not greater than 1 ppm. Also, at least one of the sample values is less than 1 ppm, so at least some of the fish are safe. O D. Yes, because it is possible that the mean is not greater than 1 ppm. Also, at least one of the sample values exceeds 1 ppm, so at least some of the fish have too much mercury.

The average gold content from an ore sample in 36 different locations is found to be 2.6 grams per ton. a. Find the 95% and 99% confidence intervals for the mean gold content in that area. Assume that the population variance is 0.09.

Suppose the population standard deviation is  = 5, an SRS of n = 100 is obtained, and the confidence level is chosen to be 99.5%. The margin of error for estimating a mean  is given by:   a. 1.576.   b. 1.165.   c. 1.404.   d. 1.223.

The manager of a fleet of automobiles is testing two brands of radial tires and assigns one tire of each brand at random to the two rear wheels of eight cars and runs the cars until the tires wear out. The data (in kilometers) follow. Find a 99% confidence interval on the difference in the mean life. Car Brand 1 Brand 2 1 36,925 34,318 2 45,300 42,280 3 36,256 35.548 4 32,100 31,950 5 37,210 38,015 6 48,360 47,800 7 38.200 37,810 8 33,500 33,215 Round your answer to 2 decimal places. Do not use commas.

One method for measuring air pollution is to measure the concentration of carbon monoxide, or CO, in the air. Suppose Nina, an environmental scientist, wishes to estimate the CO concentration in Zagreb, Croatia. She randomly selects 45 locations throughout the city measures the CO concentration at each location. Based on her 45 samples, she computes the margin of error for a 95% t-confidence interval for the mean concentration of CO in Zagreb, in µg/m³, to be 4.28. What would happen to the margin of error if Nina decreases the confidence level to 90%? Nina increases the confidence level to 99%? Nina decreases the sample size to 33 locations? Nina increases the sample size to 65 locations? Answer Bank Increase Stay the same Decrease

Twelve samples of coal from a Northern Appalachian source had an average mercury content of 0.253 ppm with a standard deviation of 0.029 ppm. Suppose the density of mercury content of coal from this source is approximately normally distributed.  a.  Find the LOWER bound of a 95% confidence interval for the mean mercury content of coal from this source. b.  Find the UPPER bound of a 95% confidence for the mean mercury content of coal from this source.

Calcium is essential to tree growth. In​ 1990, the concentration of calcium in precipitation in a certain area was 0.11milligrams per liter (mg/L).A random sample of 10 precipitation dates in 2018 results in the following data table. Complete parts​ (a) through​ (c) below. (B) With 98​% confidence, the mean concentration of calcium in precipitation in this area in 2018 is between

A ecologist wishes to estimate the mean lifespan for a specific species of bird. Assume that bird lifespan is normally distributed and that each bird's lifespan is independent of all others. The researcher analyzes a random sample of n=7 deceased birds to find out how long they lived for. It is found that the mean lifespan in the sample is y= 5 4 years with standard deviation s=13 years. Choose the correct formula for a 95% confidence interval for the true mean lifespan. X± to.975,n O xt to.975,n-1 n-1 O none of these X+ Z0.975 O x+ to.975,n-1' O xt to.975,n-15