To evaluate the effectiveness of a new type of plant food developed for tomatoes, an experiment was conducted in which a random sample of 50 seedlings was obtained from a large greenhouse having thousands of seedlings. Each of the 50 plants received 64 grams of this new type of plant food each week for 16 weeks. The number of tomatoes produced by each plant was recorded yielding the following results:
sample mean X bar = 30.35
standard deviation = 3.945
A researcher has started with a new sample and a given degree of confidence that the average number of tomatoes the seedlings produced on the new plant food is between "37.65289 and 40.00711". Suppose the sample size and standard deviation are the same as given above. What alpha did the researcher use in the construction of this statement? (Input your answer as a decimal)
I have done the following using the formula provided by in class
I isolated for z critical *
After getting z critical *, I looked at the t distributions printed table to find alpha / 2 which was within the 95% confidence interval column.
What am I doing wrong?
Your confidence interval is of the following type:
$$\bar X \pm t^*\frac{S}{\sqrt{n}}.$$
First, you need to underatand that the margin of error is half the width of the CI $(37.6589, 40.00711),$ which is $(40.00711 - 37.6589)/2 = 1.174105.$
Then the margin of error is $$t^*\frac{S}{\sqrt{n}} = t^*\frac{3.945}{\sqrt{50}} = t^*(0.5579073) = 1.174105,$$
so $t^* = 2.062.$
Now you need to find the confidence level. Sample size $n = 50$ implies degrees of freedom $n-1 = 49.$ But most t-tables won't have a row for 49 degrees of freedom, and you'll have to use the standard normal approximation instead. The value 2.062 cuts about 0.5% of the probability from the upper tail of the standard normal distribution (and also of $\mathsf{T}(49)$.) So it must be approximately a 99% CI.(To be fussy, maybe 98.8%)
Your work is a little hard to read, but now that I've done it for myself, I think your only mistake was to use the whole width of the CI in your computation instead of half the width, which is the margin of error. (It is very late here and you should check all of my computations.)