A plant breeder claims that a new variety of fruit bush he has produced gives a higher yield of fruit than the variety it will replace. How do you test the breeder's claim at a 5% level of significance?

A random sample of 10 bushes of the new variety is grown and the yields of the bushes recorded. The old variety has an average yield of 5.2kg/bush. It is assumed that the yield from each bush is an independent observation from a normal distribution.

1 Answer
Dec 8, 2017

You would use #(bar x - mu_bar x)/(sigma_bar x)# to find the z score and find the tail of tge z score.


In this problem, we need to make use of the properties of sampling distribution. We can say the mean of the sampling distribution of size 10 is equal to claimed mean of yield. The standard deviation of the sampling distribution is equal to the claimed standard deviaton divided by the square root of the sample size. You then use the z score formula of the sample mean minus the distribution mean, and then you divide by the sampling standard deviation. Then, you can use a calculator or a table to find the tail of the z score. If the tail is less than 5%, the breeder's claim is true.