Question

The built-in ldeaths data set contains measurements giving the monthly deaths from bronchitis, emphysema and asthma in the UK, 19741979, both sexes. We can easily access this data with the assignment x <- as.vector(ldeaths). 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:s:

[1] 3035 2552 2704 2554 2014 1655 1721 1524 1596 2074 2199 2512 2933 2889
[15] 2938 2497 1870 1726 1607 1545 1396 1787 2076 2837 2787 3891 3179 2011
[29] 1636 1580 1489 1300 1356 1653 2013 2823 3102 2294 2385 2444 1748 1554
[43] 1498 1361 1346 1564 1640 2293 2815 3137 2679 1969 1870 1633 1529 1366
[57] 1357 1570 1535 2491 3084 2605 2573 2143 1693 1504 1461 1354 1333 1492
[71] 1781 1915


Assume these values are a random sample from a normal population with unknown mean μ and unknown standard deviation σ.
Call the vector of observations x,using the assignment above, x <- as.vector(deaths)




g) Calculate the interquartile range of x, using R.

h) Calculate the sample variance of 4.7x.

i) Calculate an unbiased estimate of the population variance,

σ²

using this data.

j) If we assume that the population has a normal distribution, then calculate the maximum likelihood estimate of μ using this data.

k) Using this data, create an 92% confidence interval for μ, noting that the sample size is large enough so we can use a normal distribution critical value zstar.(

,

)

Answer 1

(g) 1000

(h) 371911.8

(i) 377150

(j) 2056.625

(k) The 92% confidence interval for μ is between 1928.974 and 2184.27.

The R code is:

x <- c(3035, 2552, 2704, 2554, 2014, 1655, 1721, 1524, 1596, 2074, 2199, 2512, 2933, 2889,
2938, 2497, 1870, 1726, 1607, 1545, 1396, 1787, 2076, 2837, 2787, 3891, 3179, 2011,
1636, 1580, 1489, 1300, 1356, 1653, 2013, 2823, 3102, 2294, 2385, 2444, 1748, 1554,
1498, 1361, 1346, 1564, 1640, 2293, 2815, 3137, 2679, 1969, 1870, 1633, 1529, 1366,
1357, 1570, 1535, 2491, 3084, 2605, 2573, 2143, 1693, 1504, 1461, 1354, 1333, 1492,
1781, 1915)
summary(x)
iqr = 2552 - 1552
var(x)
var(x)*(72/71)
mean(x)
t.test(x, mu = 0, conf.level = 0.92, interval = "confidence")