I am really confused by the following question:
Here is some information that I feel is relevant from part a)
- Annual rainfall totals can be approximated by a normal distribution, with a mean of 752.4mm and a standard deviation of 105.578mm.
Mean is based on data collected from 1900-1988.
The rainfall totals obtained above are a sample.
The arithmetic mean of 752.4mm is thus called a sample mean. The population mean would be calculated from an infinitely long series of annual rainfall totals. Using the sample mean, calculate a 95% confidence interval of the population mean of annual rainfall totals.
I think a t-dist. is more relevant in this instance opposed to a z dist. although I am unsure why this is the case. This is an example I have tried to follow..
These are the figures I have identified as being necessary:
Sample Mean = 752.4mm
Standard Deviation = 105.578mm
Number of Sample = 1988 – 1990 = 88
Degrees of Freedom = 88 – 1 = 87
Standard error of the mean = 105.578/[root]87 = 105.578/9.33 = 11.319
Critical t, or z value = ?
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