find the 95% confidence interval for the number of seeds for the species

Question

A botanist wishes to estimate the typical number of seeds for a certain fruit. She samples 36 specimens and counts the number of seeds in each. Use her sample results (mean = 56.4, standard deviation = 17.6) to find the 95% confidence interval for the number of seeds for the species. Enter your answer as an open-interval (i.e., parentheses) accurate to 3 decimal places. 

Answer 1

I hope you are constructing a confidence interval for the average number of seeds for this species. In other words, you are making a 95% confidence interval for the population mean. You also do NOT know what the population standard deviation is, so you will be using the "student t" distribution to calculate the "standard error". You'll need two things. First, you need the critical t value (from a t table) = t*.

1) What is your sample size (n)?

2) Degrees of freedom is needed for the t value = n - 1

3) For confidence interval, assume "two tails" with alpha = 0.05 (1 - 0.95). Look up t* _____

Since the t-table stops at 30 and then typically jumps to 60 degrees of freedom, use d.f. = 30 OR use a calculator to find the "critical t" value with the appropriate number for "degrees of freedom"

4) Calculate the standard error: SE = (sample standard deviation) divided by the square root of n

5) The margin of error (ME) = t* x SE <-- this really tells you "how wide" your estimate needs to be!!

6) Using interval notation (left, right) calculate these (mean - ME, mean + ME)

Hope this helps!

(By the way, some might suggest that you can use the z table because your sample size is > 30. You are only justified in using z if the sample size is MUCH larger, *or* you need to know the population standard deviation.)