Question

The built-in R dataset swiss gives Standardized fertility measure and socio-economic indicators for each of 47 French-speaking provinces of Switzerland at about 1888. The dataset is a data frame containing 6 columns (variables). The column Infant.Mortality represents the average number of live births who live less than 1 year over a 3-year period. We are interested in the Infant.Mortality column. We can convert the data in this colun to an ordinary vector x by making the assignment x <- swiss$Infant.Mortality. Then we can easily access the data. We can also get the data by entering the values 1 by 1. (You would be wise to not do the 1 by 1 entry.) The following is a screen print of the data values: [1] 22.2, 22.2, 20.2, 20.3, 20.6, 26.6, 23.6, 24.9, 21.0, 24.4, 24.5, 16.5, 19.1, 22.7 [15] 18.7, 21.2, 20.0, 20.2, 10.8, 20.0, 18.0, 22.4, 16.7, 15.3, 21.0, 23.8, 18.0, 16.3 [29] 20.9, 22.5, 15.1, 19.8, 18.3, 19.4, 20.2, 17.8, 16.3, 18.1, 20.3, 20.5, 18.9,, 23.0 [43] 20.0, 19.5, 18.0, 18.2, 19.3 l) Using this data, create a 99% prediction interval for μ, noting that the sample size is large enough so we can use a normal distribution critical value zstar. ( , ) m) Using this data, we create a 1% level test of H0: μ=21 versus the alternative Ha: μ < 21. We will reject H0 if z = x − 21 s 47 < zstar where s is the sample standard deviation. What is the value of zstar? (Calculate from normal distribution) n) Continuing from part m, what is the value of z? o) Continuing from parts m and n, what is the p value of the test.

Answer 1