An experiment was conducted to determine the effect of a position in a garden on the heights of sunflowers. Two samples of heights were obtained: One of sunflowers growing in a shady part of the...
An experiment was conducted to determine the effect of a position in a garden on the heights of sunflowers. Two samples of heights were obtained: One of sunflowers growing in a shady part of the garden, and one of sunflowers growing in a sunny part of the garden. The data (in meters) are in the attached JPG file.
Q) Carry out a sample z-test to investigate whether there is a difference between the mean height of sunflowers growing in a shady part of a garden and the mean height of sunflowers growing in a sunny part of a garden. Include in answer a statement of the null and alternative hypothesis, including an explanation of the meaning of any symbols that you use(other than H0 and H1) the value of the test statistic your conclusions
For the plants in shade we have `bar(X)_1=2.88,sigma_1=.512`
For the plants in the sun we have `bar(X)_2=3.62,sigma_2=.878`
(`bar(X)` is the sample mean and `sigma` is the population standard deviation.)
1. `H_0:` `mu_1=mu_2` (the population means are the same)
`H_1:` `mu_1 != mu_2` (the population means differ)
2. Suppose we want to be 95% certain. Then `alpha=.05` and `alpha/2=.025` . Then from a standard normal table we find the critical value to be `z=+-1.96` . (The critical regions will be z<-1.96 or z>1.96)
3. To compute the test value we use `z=((bar(X)_1-bar(X)_2)-(mu_1-mu_2))/sqrt((sigma_1^2)/n_1+(sigma_2^2)/n_2)`
Note that from the null hypothesis `mu_1-mu_2=0` .
`n_1=30,n_2=28` as these were the number of test subjects in the samples.
4. The test value is in the critical region so we reject the null hypothesis.
5. There is sufficient evidence to claim that the means in the heights of the plants differ.
Using a TI-83 I got z=-3.893 and `p=9.9"x"10^(-5)` which indicates to reject the null hypothesis.