Question

The high temperatures (in degrees Fahrenheit) of a random sample of 7 small towns are: Assume high temperatures are normally distributed, Based on this data, find the 99% confidence interval of the mean high temperature of towns. Enter your answer as an open-interval (i.e., parentheses) accurate to two decimal places (because the sample data are reported accurate to one decimal place). 998 C.1. = Answer shoutd be obtained without any preliminary rounding. However, the criticat value may be rounded 103 decimal places. SAT scores are normally distributed with a mean of 1,500 and a standard deviation of 300. An administrator at a college is interested in estimating the average SAT score of first-year students. If the administrator would like to limit the margin of error of the 82% confidence interval to 5 points, how many students should the administrator sample? Make sure to give a whote number answer.

Answer 1

The question involves calculating confidence intervals and necessary sample size for a known margin of error. The 99% confidence interval requires the sample mean and standard deviation, which are not provided. The sample size for limiting SAT score margin of error to 5 points can be calculated using the margin of error formula with a given confidence level and standard deviation.

Step-by-step explanation:

The student's question relates to finding confidence intervals for given data sets and determining the necessary sample size to achieve a specific margin of error for a confidence interval. When constructing confidence intervals for normally distributed data, where the population standard deviation is unknown, we use a t-distribution. With a known population standard deviation and a normal distribution, we use a z-score to calculate the confidence interval.

To find a 99% confidence interval for the mean high temperature of towns using the provided data, we would typically need the sample mean and the sample standard deviation of those temperatures. Unfortunately, the student hasn't provided these values, therefore we're unable to calculate this confidence interval.

For the SAT scores, to limit the margin of error to 5 points at an 82% confidence level, we can use the formula for margin of error:

ME = z*(σ/√n), where ME is the margin of error, z is the z-score corresponding to the confidence level, σ is the population standard deviation, and n is the sample size. Re-arranging for n and substituting the given values will allow us to calculate the necessary sample size.