Question

The state of Oregon would like to test whether the average GPA of college students is different than 3.2 or not for scholarship purposes. They take a random sample of 1000 college students in Oregon. Here is the summary statistics.

Sample mean, x ¯	3.18
Sample size, n	10,000
Sample standard deviation, s	0.46

Refer to the information in the previous problem.

Suppose we test like to test the hypotheses,

H₀ : μ = 3.2  vs.  H_a: μ ≠ 3.2

with a significance level of α = 0.05. Which of the following methods are a valid way to test the hypotheses?

For each of the following methods indicate whether they are a valid way to test the hypotheses. In other words, is the method described an appropriate way of finding a p-value and trusting the conclusion is a correct conclusion considering the data.

Method 1. Using our experience on the subject, we try to guess what the p-value is likely to be.  

Method 1 is                            [ Select ]                       ["a valid", "NOT a valid"]           procedure.

Method 2. Take 10000 samples from the population of all Oregon college students, for each sample record the mean. Find the p-value by determining the proportion of these means that are greater than different than 3.2.

Method 2 is                            [ Select ]                       ["a valid", "NOT a valid"]           procedure.

Method 3. Calculate the 95% confidence interval for μ , if the null hypothesized value does not fall within the interval then we can reject the null hypothesis.

Method 3 is                            [ Select ]                       ["a valid", "NOT a valid"]           procedure.

Method 4. Perform a one sample z-test for the proportion of GPAs greater than 3.2

Method 4 is                            [ Select ]                       ["a valid", "NOT a valid"]           procedure.

Method 5. To solve for the p-value, they can calculate a t-test statistic and degrees of freedom and find the area under the t-distribution curve that is more unusual than the calculated test statistic.

Method 5 is                            [ Select ]                       ["a valid", "NOT a valid"]           procedure.

Answer 1

1.Method 1. Using our experience on the subject, we try to guess what the p-value is likely to be.  

it is not a valid estimate since we need some statistical evidence like p-value,level of significance etc to reach at a conclusion,simply guessing isnt a good idea.

2.Method 2. Take 10000 samples from the population of all Oregon college students, for each sample record the mean. Find the p-value by determining the proportion of these means that are greater than different than 3.2.

it is not a valid estimate since we have the total sample size to be 10000 and taking out of 10000 samples isnt possible also it will then be a sampling distribution and we are given the information regarding only 1 sample.

3.Method 3. Calculate the 95% confidence interval for μ , if the null hypothesized value does not fall within the interval then we can reject the null hypothesis.

it is not a valid estimate since we cant judge the value of null hypthesis based on the confidence interval to reach out a conclusion about acceptance or rejection of null hypothesis.

4.Method 4. Perform a one sample z-test for the proportion of GPAs greater than 3.2

it is not a valid estimate since we are given the mean and not the proportions as a result we cant apply z-test for proportions.

5.Method 5. To solve for the p-value, they can calculate a t-test statistic and degrees of freedom and find the area under the t-distribution curve that is more unusual than the calculated test statistic.

it is not a valid estimate since the sample size is 10000 which is large so we can apply large sample test,but t-test is not a large sample test hence we cant make conclusions using t-test.