Question

Create and solve a population mean problem that you may be interested in. Gather appropriate data and post your problem. Include the null and alternative hypothesis, alpha value, p-value, and a conclusion. Make sure that you use appropriate terminology, specify whether you are using the classical method or the p-value method, and fully explain your solution.

How do we determine that we are to fail to reject the null hypothesis? What process do we use to determine this result?

A principle is curious about how many hours per week students’ study in the grade 10 psychology class. It is believed for each student to be successful in the class, they must spend at least 6 hours/week studying. He went on to ask 7 random students to keep track of the amount of time each week they study and it was concluded that the average was 6.5 hrs.

Results: 6.5 9 7 7 8 5 3

Standard deviation: 2 hours

H0: μ = 6 hrs HA: μ > 6 hrs

α = 0.10 significance level

N= 7 X= 6.5

Answer 1

Any population mean problem first means that we state the Null and the Alternate Hypothesis.

In the question given above:

The Null hypothesis is H0 :

Here it means that the population mean i.e. the entire population is expected to study at least 6 hours per week studying.

Hence Null Hypothesis is known as the given truth or the truth that we want to test.

Alternative Hypothesis is H_a:

This is the statement that we think might be true and what we need to test.

The value of significance i.e. alpha is our level of significance.

The standard deviation of the sample here is known i.e. 2 hours

Which means s = 2 hours
But N =7 which is <30

Hence, we these two conditions tell us that we shall use the t-statistic instead of the z-statistic as the appropriate test statistic.

Here,

s = 2 hours

Hence,

t = 0.6614

Since level of significance is 0.10, we find the critical value corresponding to it.

This is found either by the probability distribution table which gives the probability value of 0.10.

We get that value as -1.43976

The value that we computed as the t-statistic is 0.6614

Since, t> critical value, we cannot the reject the null hypothesis

Had the t-statistic been less than -1.439765, we would have rejected the null hypothesis.

Hence, as we cannot reject the null hypothesis, it means that the principal's hunch that students' average study time per week is close to 6 hours is true.

This is how the entire test is conducted.