Kierra recorded her daily exercise activity for a month in this frequency table. Which problem shows how to solve for the follow
1 answer:
You might be interested in
Answer:
As the pvalue of the test is 0.02097 < 0.1, using a .10 significance level, he can conclude that the average trunk diameter of a pine tree is greater than the average diameter of a spruce tree.
Step-by-step explanation:
Using a .10 significance level, can he conclude that the average trunk diameter of a pine tree is greater than the average diameter of a spruce tree
The null hypothesis is that they are equal(that is, subtraction between the means is 0), while the alternate hypothesis is that they are greater(that is, subtraction between the means is larger than 0). So
In which is the mean pine trees height while
is the mean spruce trees height.
The test statistic is:
In which X is the sample mean, is the value tested at the null hypothesis and s is the standard error.
0 is tested at the null hypothesis:
This means that
Mean trunk diameter (cm) 35 30:
Pine trees - 35
Spruce trees - 30
The sample mean is given by the subtraction of the means. So
Sample Size 40 80
Population variance 160 160
This means that the standard error for each sample is given by:
The standard error of the difference is the square root of the sum squared of the standard error of each sample.
Test statistic:
Pvalue of the test:
Probability of a sample mean larger than 5, which is 1 subtracted by the pvalue of Z when X = 5.
Looking at the z-table, z = 2.04 has a pvalue of 0.9793.
1 - 0.9793 = 0.0207
As the pvalue of the test is 0.02097 < 0.1, using a .10 significance level, he can conclude that the average trunk diameter of a pine tree is greater than the average diameter of a spruce tree.