Amedical study showed that girls under sixteen who are treated for hodgkin’s disease face an exceptionally high risk of developing a second type of cancer later in life (new york times, 3/21/96). the study followed 1,380 girls after their original treatment, and 88 of them, or 0.0638, developed second cancers. only four of them, or 0.0029, would have been expected based on data for the population at large. test the null hypothesis that girls treated for hodgkin’s disease are not at increased risk for a second cancer (h0: p ≤ 0.0029) against the alternative hypothesis that they are at increased risk (ha: p > 0.0029). use a 10% level of significance. what are the test statistic and its sampling distribution?

According to a research study, parents with young children slept 6.4 hours each night last year, on average. a random sample of 18 parents with young children was surveyed and the mean amount of time per night each parent slept was 6.8. this data has a sample standard deviation of 0.9. (assume that the scores are normally distributed.) researchers conduct a one-mean hypothesis at the 5% significance level, to test if the mean amount of time parents with young children sleep per night is greater than the mean amount of time last year. the null and alternative hypotheses are h0: μ=6.4 and ha: μ> 6.4, which is a right-tailed test. the test statistic is determined to be t0=1.89 using the partial t-table below, determine the critical value(s). if there is only one critical value, leave the second answer box blank.