Discuss the below:
1. An election poll reported that a candidate had an approval rating of 45% with a margin of error E of 4%. Construct a confidence interval for the proportion of adults who approve of the candidate.
2. You are given the sample mean and the sample standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Which interval is wider? If convenient, use technology to construct the confidence intervals.
A random sample of 48 eight-ounce servings of different juice drinks has a mean of 97.3 calories and a standard deviation of 47.6 calories.
The 90% confidence interval is
The 95% confidence interval is
Which interval is wider?
3. The state test scores for 12 randomly selected high school seniors are shown on the right.
1422 1224 989
696 729 839
725 749 540
627 1441 940
Assume the population is normally distributed.
(a) Find the sample mean.
(b) Find the sample standard deviation.
(c) Construct a 99% confidence interval for the population mean µ.
A 99% confidence interval for the population mean is ______
4. A researcher wishes to estimate, with 95% confidence, the proportion of adults who have high-speed Internet access. Her estimate must be accurate within 3% of the true proportion.
(a) Find the minimum sample size needed, using a prior study that found that 56% of the respondents said they have high-speed Internet access?
(b) What is the minimum sample size needed assuming that no preliminary estimate is available?
5. People were polled on how many books they read the previous year. How many subjects are needed to estimate the number of books read the previous year within one book with 90% confidence? Initial survey results indicate that σ = 18.6 books.
A 90% confidence level requires __________ subjects.
6. Use the confidence interval to find the estimated margin of error. Then find the sample mean.
A biologist reports a confidence interval of (1.9, 3.3) when estimating the mean height (in centimeters) of a sample of seedlings.
The estimated margin of error is
The sample mean is