Question

To evaluate the effectiveness of a new type of plant food developed for tomatoes, an experiment was conducted in which a random sample of 56 seedlings was obtained from a large greenhouse having thousands of seedlings. Each of the 56 plants received 80.5 grams of this new type of plant food each week for 9 weeks. The number of tomatoes produced by each plant was recorded yielding the following results: ?bar =31.08 . ?=4.325

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 "35.64258 and 37.63742". 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?

Answer 1

Here given ,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 "35.64258 and 37.63742.

that is we consider confidence interval for new plant food is ( 35.64258 and 37.63742 ).

Akso given ,

Lower limit = Sample mean - Margin of error . = 35.64258

Upper limit = Sample mean + Margin of error = 37.63742

To find margin of error = upper limit - lower limit / 2 = 37.63742-35.64258 / 2 =1.9984 /2 = 0.9974.

Formula for margin of error E as,

    

  

From this tc = critcal value at df = 56-1=55 using two tailed area .

In excel use command as =TDIST(x value, df , tails) so here command is as ,

=TDIST(0.24,55,2) then hit enter you will get probability that is alpha as 0.81

So here we say that, alpha =0.81 the researcher use in the construction of this statement