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 = 30.35

s = 3.945

(a) Assuming that the seedlings chosen are taken from a population which is normally distributed, determine a 95% confidence interval estimate for the average number of tomatoes that would have been produced by all the seedlings in the greenhouse if they have received 64 grams of the new plant food, once a week for 16 weeks. Use three decimals.

n = random sample

n = 50

sample mean = 30.35

s = 3.945

alpha = 0.05

95% Confidence Interval = z*

                    = z alpha/2 = 0.025

                    = 1.96

Margin of Error = z* standard deviation / square root of n

            = 1.96 * (3.945 / square root of 50)

            = 1.093498211

LL = sample mean x - Margin of Error

= 30.35 - 1.96 * 1.093498211

= 28.10674351

UL = sample mean x + Margin of Error

= 30.35 + 1.96 * 1.093498211

= 28.20674351

Are these the correct solutions and answers? If not, what are the correct solutions and answers?

I am thinking that maybe the 64 grams from the description above and once every 16 weeks changes everything completely, but I am not sure.

  • 1
    $\begingroup$ Just a comment, not sure if it helps. Reading your question, it is unclear to me what is meant by "Assuming that the seedlings chosen are taken from a population which is normally distributed". Over what parameter are the population normally distributed? Is it implied that the number of tomatoes produced if given the new type of food is normal? $\endgroup$ – them Mar 15 '17 at 3:15
  • 1
    $\begingroup$ It looks like you've done pretty much everything right for the purposes of the problem, though I have not checked the numbers in detail. The one thing is that technically you should use a t distribution on $49$ degrees of freedom to get your critical value instead of the normal distribution. The sample size is big enough here that there's not too much difference. It would be $2.01$ rather than $1.96.$ It may be that you haven't done t statistics in class yet and they want you to use the normal as an approximation, in which case, that's fine. $\endgroup$ – spaceisdarkgreen Mar 15 '17 at 3:35

Checking to be sure, using R statistical software, mostly as a calculator and instead of a t table.

Your method is pretty much on the right track, but it seems you have included the factor 1.96 twice in finding LL and then made a typographical error for UL. (Notice that $LL$ will always be below $\bar X$ and $UL$ will always be above.)

The style of CI you want is

$$\bar X \pm t^*\frac{S}{\sqrt{n}},$$

where $t^* = 2.01$ cuts 2.5% of the area from the upper tail of Student's t distribution with $n - 1 = 50 - 1 = 49$ degrees of freedom (as per @spaceisdarkgreen's Comment).

qt(.975, 49)
## 2.009575

Then the 95% CI is $(29.229,\, 31.471).$

pm = c(-1,1);  30.35 + pm*2.01*3.945/sqrt(50)
## 29.22861 31.47139

If you use $z^* = 1.96$ from the standard normal distribution as an approximation, you get $(29.257,\, 31.444)$ instead.

30.35 + pm*1.96*3.945/sqrt(50)
## 29.2565 31.4435

One rule for the number of decimal places to show in a confidence interval is 'one more than in the original data'. But you don't have the original data, so I used one more than given for the sample mean. Also, showing three places illustrates the (slight) difference between the exact and approximate methods.

Strictly speaking, you should always use Student's t distribution when the population standard deviation $\sigma$ is unknown and estimated by the sample standard deviation $S,$ regardless of the sample size. One reason for this is that the sample size at which $t^* \approx z^*$ depends on the confidence level. For a 95% CI it's around $n = 30;$ for a 99% CI it's more like $n = 70.$

[However, if you haven't covered the t distribution yet, or if you were instructed to use the standard normal approximation, then the choice is out of your hands.]

Finally, before you submit this, please double-check to make sure I didn't make typographical errors. I'm really good at finding them in other people's work, not so good with my own.


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