Question 23 6 pts Social science researchers used Facebook posts from users in the US to analyze "profanity use rates" by collecting large samples of words from users status updates and computing the proportion of swear words in the sample of words. Suppose that for their sample of 100,000 words from status updates in California, the sample proportion of profanity use was 0.0044. Suppose also for their sample of 100,000 words from status updates in New York, the sample proportion of profanity use was 0.0046. The software output to test if these different proportions represent a real difference or if this is just random chance is below. Use it to answer the questions that follow. Proportion Proportion Standardized P-Value CA NY Difference Obs Stat 0.0044 0.0046 -0.0002 -0.66817 0.50402 1. State the null and alternative hypotheses for this test 2. What is the point estimate is equal to? How was it computed? 3. What is the test statistic is equal to? 4. What distribution does the test statistic follow? 5. What is the conclusion of the hypothesis test? Use a significance level of 0.01 and state your conclusion in terms of the problem. In other words, what does the evidence say about swearing on Facebook for the two states? 6. Based on your answer above, would you expect 0 to be in a 99% confidence interval for the difference of two proportions? Why