Since your sample size is small and the problem doesn't say that the population has a normal distribution, you may want to use a t-table to determine the 95% confidence interval. A general example of a formula is: CI95 = mean + or - (t-value)(sd/√n)
Note: sd = standard deviation.
You will need to determine mean and standard deviation before you can plug the values into the formula to determine the interval. The sample size is 6.
For the second part, you will need a formula to determine sample size. There are different formulas you can use; here is an example: n = [(t-value) * sd)/E]^2
Note: E = 5
I hope this will help.
Bacterial samples were taken from several surfaces in a daycare and incubated in petri dishes. After that the number of bateria colonies were counted, with the following results 412, 437, 425, 476, 451, 483.
1) construct a 95% confidence interval for the mean number of colonies
2) after constructing the interval above, the resarcher notices that it is quite wide, and wishes to do a follow-up study with greater precision. How large a sample is reuired to produce a 95% confidence interval with margin of error 5?
IM SO CONFUSED!!
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One other comment: if you are expected to use a z-value instead of a t-value, the z-value would be 1.96 using a z-table for a normal distribution.
1 answer