90% = mean ± 1.645 SEm
SEm = SD/√n
I used the table in the back of my statistics text labeled "areas under normal distribution" to find the proportion/probability (±5%) to get Z = 1.645. I assume that you have a similar table available.
I have like 9 more of these problems, can someone please help me out so I can get started. thanx JS
In a random sample of 35 tractors, the annual cost of maintenance was $4,425 and the standard deviation was $775.
Construct a 90% confidence interval for this.
Assume the annual maintenance costs are normally distributed.
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