Question

1) Suppose a researcher is interested in understanding the variation in the price of store brand milk. A random sample of 36 grocery stores selected from a population and the mean price of store brand milk is calculated. The sample mean is $3.13 with a population standard deviation of $0.23. Construct a 92% confidence interval to estimate the population mean.

2) Construct 90%, 95%, and 99% confidence intervals to estimate μfrom the following data. State the point estimate. Assume the data come from a normally distributed population.

12.7 11.6 11.9 11.9 12.5

11.4 12.0 11.7 11.8 12.7

3) Is the environment a major issue with Americans? To answer that question, a researcher conducts a survey of 1255 randomly selected Americans. Suppose 690 of the sampled people replied that the environment is a major issue with them. Construct a 95% confidence interval to estimate the proportion of Americans who feel that the environment is a major issue with them. What is the point estimate of this proportion?

Appendix A Statistical Tables

(Round the intermediate values to 3 decimal places. Round your answer to 3 decimal places.)

___ ≤ p ≤ ___

The point estimate is ___

Answer 1

Confidence Interval: 3.13 +0.0671 (+2.1%) [3.063 – 3.197] Error Bar: Steps: C1 = X + 2x = 3.13 + 1.7507938 = 3.13 + 0.0671

Confidence Interval: 12.02 +0.239 (+2.0%) (11.781 – 12.259] Error Bar: Steps: C1 = X + Zx = 12.02 + 1.6449x0.4500 2) 90% = 12.02 + 0.239

Confidence Interval: 12.02 +0.284 (+2.4%) (11.736 - 12.304] Error Bar: Steps: CI = X + 2x = 12.02 + 1.9600x = 1202-10000, 0.4590 *710 = 12.02 + 0.284 2) 95%

Confidence Interval: 12.02 0.374 (3.1%) [11.646 – 12.394] Error Bar: Steps: CI = X + 2x Vn 0.4590 = 12.02 + 2.5758x- 710 2) 99% = 12.02 + 0.374

Proportion of positive results = P = x/N = 0.550 Lower bound = 0.522 Upper bound = 0.578 3.

Probability of success on a single trial 0.55 Number of trials 1255 Number of successes (x) 690 Binomial probability: P(X = x) 0.02262829327
Its a confidence interval .pl?☺️