Among 1815 randomly selected high school students surveyed by the Centers for Diseases Control, 300 said they were current smokers. d. Interpret the 95% confidence interval in this context.

In terms of z-scores, a 95% confidence interval consists of which of the following ranges?        A. μ ± 1.96(z)       B. μ ± 2.05(z)       C. μ ± 1(z)       D. μ ± 2.58(z)

An insurance company selected samples of clients under 18 years of age and over 18 and recorded the number of accidents they had in the previous year. We are interested in determining if the accident proportions differ between the two age groups. The results are shown below. a =0.05 Under age of 18 Over age of 18 n₂ =500 n₁ = 480 Number of accidents = 160 Number of accidents = 140 The test statistic is A 90% confidence interval for the proportion between two age groups is approximately Given a 5% level of significance, what decision should be reached about the null hypothesis

QUESTION 1 Calculate a 95% confidence interval: Of the 1220 respondents for whom we have information, 178 said that they had been divorced at least once.

A sample of n = 200 college students are asked if they believe in extraterrestrial life and 120 of these students say that they do. The data are used to test if more than half of college students believe in extraterrestrial life. d. Find a 95% confidence interval   e. Write a conclusion for the confidence interval

To test the effect of a physical fitness course on one's physical ability, the number of sit-ups that a person could do in one minute, both before and after the course, was recorded. Ten individuals are randomly selected to participate in the course. The results are displayed in the following table. Using this data, find the 90%90% confidence interval for the true difference in the number of sit-ups each person can do before and after the course. Assume that the numbers of sit-ups are normally distributed for the population both before and after completing the course. Sit-ups before 2929 3636 2828 5050 2323 3737 3232 2929 4040 2121 Sit-ups after 4545 3838 3535 5656 4141 5555 4747 4545 4444 2626 Copy Data Step 1 of 4:  Find the point estimate for the population mean of the paired differences. Let x1x1 be the number of sit-ups before taking the course and x2x2 be the number of sit-ups after taking the course and use the formula d=x2−x1d=x2−x1 to calculate the paired…

A survey of 2450 golfers showed that 281 of them are left-handed. Construct a 90% confidence interval for the proportion of golfers that are left-handed. Question 10 options: (0.104, 0.125) (0.369, 0.451) (0.100, 0.130) (0.203, 0.293)

If X=101, o=27, and n= 32, construct a 95% confidence interval estimate of the population mean, p

A 95% confidence interval for a population proportion, p, is 0.23 < p < 0.27. What was the sample proportion?

50 of the 600 medical students who were chosen at random stated they intended to work in a rural area. Calculate the 90% confidence interval for the actual percentage of medical students who intend to practice in rural areas. The responses shown here are estimates.