This is statistics course, sampling distributions, estimation and tests of significance units.I need help with this. I attached formula sheet in case you need it.

II. Probability P(AUB) = P(A) + P(B) - P(An B) P(AB) = P(An B) P(B) E(X) = Ux = Expi Var(X) = 0; = _(x - ux)2pi If X has a binomial distribution with parameters n and p, then: P(X = k) = k) p* (1 - p)" -* HI = np Or = Vnp(1 - p) Ho = P Op = P(1- p n If x is the mean of a random sample of size n from an infinite population with mean u and standard deviation O, then: MY = HIII. Inferential Statistics Standardized test statistic: statistic - parameter standard deviation of statistic Confidence interval: statistic + (critical value) . (standard deviation of statistic) Single-Sample Statistic Standard Deviation of Statistic Sample Mean Sample Proportion p(1 - p) n Two-Sample Statistic Standard Deviation of Statistic Difference of sample means 722 Special case when 61 =02 ol+1 n2 Difference of PI (1 - PI) + P2 (1 - P2) sample proportions n1 n2 Special case when P1 = P2 Jp(1 - P) 1 + 1 Vn, n2 Chi-square test statistic = \\ (observed - expected) expectedD DIRECTIONS: Use this information for questions 13 to 15. A fair, six-sided number cube is rolled 60 times and the results are shown in the table below. Number showing: 1 2 3 5 6 Frequency: 12 13 11 8 6 10 D Question 13 1 pts Compute the chi-squared test statistic. O 0.567 O 1.20 O 3.40 O 3.75 O 4.28D Question 14 1 pts Which of these is most appropriate test for the situation above? Test for a difference of means One-sample t-test O Goodness-of-fit test Test of independence Test of homogeneity D Question 15 1 pts Given the hypothesis that the cube is fair, and the alternative hypothesis that the cube is not fair, conduct the test at a significance level of a = 0.10. The data is a good fit because the p-value is above 0.10. The data is a good fit because the p-value is below 0.10. The data is a good fit because the p-value and a are approximately equal. The data is a not a good fit because the p-value is above 0.10. O The data is a not a good fit because the p-value is below 0.10.\f