Is it snowing where u are?

On a certain portion of an experiment, a statistical test result yielded a p-value of 0.18. what can you conclude? 2(0.18) = 0.36 < 0.5; the test is not statistically significant. if the null hypothesis is true, one could expect to get a test statistic at least as extreme as that observed 18% of the time, so the test is not statistically significant. if the null hypothesis is true, one could expect to get a test statistic at least as extreme as that observed 82% of the time, so the test is not statistically significant. 0.18 > 0.05; the test is statistically significant. p = 1 - 0.18 = 0.82 > 0.05; the test is statistically significant.

And intervening opportunity is an example of

In a recent telephone survey, 5,000 randomly selected teenagers were asked to cite their primary social network site. six of 10 teenagers said they use mybook as their primary social network site. a 95% confidence interval to estimate the true proportion of teenagers who use mybook as their primary social network site is found to be (0.2964, 0.9036). which of the following is a correct interpretation of the confidence level? ninety-five percent of all samples of this size would yield a confidence interval of (0.2964, 0.9036). there is a 95% chance that the true proportion of teenagers who use mybook as their primary social network site is in the interval (0.2964, 0.9036). ninety-five percent of the time, the procedure used to generate this interval will capture the true proportion of teenagers who use mybook as their primary social network site. ninety-five percent of all the samples of size 5,000 lie in the confidence interval (0.2964, 0.9036).