A study was commissioned to find the mean weight of the residents in certain town. The study examined a random sample of 127 residents and found the mean weight to be 151 pounds with a standard deviation of 21 pounds. At the 95% confidence level, use the normal distribution/empirical rule to estimate the margin of error for the mean, rounding to the nearest tenth.
Question
A study was commissioned to find the mean weight of the residents in certain town. The study examined a random sample of 127 residents and found the mean weight to be 151 pounds with a standard deviation of 21 pounds. At the 95% confidence level, use the normal distribution/empirical rule to estimate the margin of error for the mean, rounding to the nearest tenth.
Solution
To calculate the margin of error for the mean at the 95% confidence level, we will use the formula for the margin of error (E) which is E = Z * (σ/√n), where Z is the Z-score, σ is the standard deviation, and n is the sample size.
-
Identify the Z-score: For a 95% confidence level, the Z-score is approximately 1.96 (this value represents the number of standard deviations away from the mean that encompasses 95% of the data in a normal distribution).
-
Identify the standard deviation (σ): From the problem, we know that the standard deviation is 21 pounds.
-
Identify the sample size (n): From the problem, we know that the sample size is 127 residents.
-
Substitute these values into the formula: E = 1.96 * (21/√127)
-
Calculate the square root of 127, which is approximately 11.27.
-
Divide the standard deviation by the square root of the sample size: 21/11.27 = 1.86.
-
Multiply this result by the Z-score to find the margin of error: 1.96 * 1.86 = 3.65.
-
Round to the nearest tenth: 3.7 pounds.
So, at the 95% confidence level, the margin of error for the mean weight of the residents in the town is approximately 3.7 pounds.
Similar Questions
A study was commissioned to find the mean weight of the residents in certain town. The study found a confidence interval for the mean weight to be between 156 pounds and 176 pounds. What is the margin of error on the survey?
A study was commissioned to find the mean weight of the residents in certain town. The study found the mean weight to be 179 pounds with a margin of error of 10 pounds. Which of the following is a reasonable value for the true mean weight of the residents of the town?
Weights of women in one age group are normally distributed with a standard deviation of 10.43 kg. A researcher wishes to estimate the mean weight of all women in this age group.Find how large a sample must be drawn in order to be 95% confident (use z=2) that the sample mean will not differ from the population mean by more than 1.30 kg.
A random sample of 12 items is taken and is found to have a meanweight of 50 grams and a standard deviation of 9 gramsWhat is the mean weight of population1. with 95% confidence2. with 99% confidence
An investigator wants to assess whether the mean (mu) = the average weight of passengers flying on small planes exceeds the FAA guideline of average total weight of 185 pounds (passenger weight including shoes, clothes, and carry-on). Suppose that a random sample of 25 passengers showed an average total weight of 200 pounds with a sample standard deviation of 59.5 pounds. Assume that passenger total weights are normally distributed.What is the value of the test statistic? Group of answer choicest = 1.26t = 1.65t = 1,8t = 1.50
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.