lo Sample Size 900 Response of interest Count for Response Sample Proportion es Yes es 488 es lo es Confidence Coefficient 0.95

This research was conducted with Korean adolescents in the Los Angeles area. Would you be willing to generalize the results of this study to Korean adolescents who live in other regions of the country? Why or why not.

Pedro was working on a report on Swiss and Austrian psychiatrists. He decided to find a 90 percent confidence interval for the difference in mean age at the time of significant psychiatric discoveries for Swiss versus Austrian psychiatrists. He found the ages at the time of psychological discovery of all the members of both groups and found the 90 percent confidence interval based on a t-distribution using a calculator. The procedure he used is not appropriate in this context because O The sample sizes for the two groups are not equal. O The entire population is measured in both cases, so the actual difference in means can be computed and a confidence interval should not be used. O Age at the time of psychological discovery occurs at different intervals in the two countries, so the distribution of ages cannot be the same. O Age at the time of psychological discovery is likely to be skewed rather than bell shaped, so the assumptions for using this confidence interval formula are not…

hi,  need steps and answers pls.

The following contingency table records some information about a sample of visitors to Old Fort William over a long weekend in August. Local Other Canadian Non-Canadian Total Young Child 40 75 35 150 Teen 25 50 20 95 Young Adult Mid Aged Adult 140 85 125 200 465 100 195 380 Senior Adult 120 165 125 410 Total 410 685 405 1500 Given the above information and choosing a level of significance of a = 0.01; you should reject the null hypothesis that there is at least one difference in origins among age groups that is statistically significant. True False

You are trying to study the level of personal net worth of a low income populaon. In parcular you are interested in find the confidence interval of the mean level of net worth (net worth can be posive or negave and is measured in thousands of dollars). You do not know the distribuon of the level of net worth of the populaon. For your study, you collected a sample from 100 persons asking them their level of net worth. The results were such that the sample meanXˉ = 3 and the sample variance Sn2 = 100 (Assume you choose the persons randomly, they report their true net worth, and the populaon is sufficiently large.) From this data, find the 90% confidence interval of the mean level of net worth.(Do not use t distribuon)

please use the calculator method to answer a ti83 or ti84 thank you    Of 400 licensed drivers selected at random, 52 are problem drinkers. Construct a 90% “plus four” confidence interval for the true proportion of drivers who are problem drinkers. What is the margin of error in this confidence interval?

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how do you develop 95% confidence intervals for the mean age and house income of subcsribers in excel? This could be any number, I just don't know how to input something like this on excel