Apply: Signature Assignment: Globalization and Information Research

Competency 2 Part 2 Hypothesis testing:

Part 2: First Hypothesis Test

This is a one-tail test in which the population is large enough to use the “z score” as the test statistic. However, because we want to ensure minimal errors, we will be comparing the p-value with the area of significance (0.05). In order to compute the p-value, we need to first identify the z-score. This test calls for a 90% confidence level which means that the z score is 1.645.

Compute these two values:

(1) Identify the z-score

(2) Identify the Ho and the H1. µ is the symbol for “sample population”.

The null and alternate hypotheses are defined by seconds so we are using 150 seconds in the hypothesis statements for this first test:

Ho : µ ≥ 150;

H1 : µ < 150

You will be using the z score to compute the p-value. In this case, we are conducting a one-tail, lower test . . . the z-score for a 90% confidence level is 1.645. Because this is a lower tail test (left hand test), the z-value that you will use to obtain the p-value is: 1.645.

(3) Conduct the hypothesis test: (compute the p-value to compare with the area of significance: .05)

You may use the p-value calculator (below) to calculate the p-value statistic. Keep in mind that the p-value is not the P (probability) but the actual statistic that is used to determine the outcome when the p-value is compared to the area of significance. You will be computing “p from z” with a z score of 1.645.

https://goodcalculators.com/p-value-calculator/

(4) Identify your Analysis: For this test, we are comparing the p-value to the area of significance (0.05). Where does the p-value fall? What does that mean when we apply the decision rule? Now, explain what this tells you, i.e. what are the results of the test, and what would you recommend.

Hypothesis Test: Part 2

This is a test of means for two independent samples with unknown variances. The samples are assumed unequal, however, because the sample size is large enough, we can assume that the data distribution operates as a normal distribution. Sample 1 (represented by µ1) is the data from the PT (traditional) protocol. Sample 2 (represented by µ2) is the data from the PE (new) protocol. We are conducting this test at the 90% confidence level so the z score is: 1.645. Since this is a left tail test, the z score is a positive value.

We want to test whether the mean ST with protocol PE is smaller than the mean ST with protocol PT.

The PE value comes from subtracting PT from ST. Therefore, ST minus PT equals PE. Let me know if you have questions about that. See the null and alternate hypothesis below. Because I have given you the critical z/t value below, you will not need to compute these values—you can skip that step).

1. The null and alternate hypotheses are: (Remember: µ1 = PT and µ2 = PE)

Ho: µ2 > µ1;

H1: µ2 < µ1.

This is a one-tailed (left hand) test. Because of the very large samples, there is no real difference between finding the critical value with a normal distribution (z score) or the Student’s t distribution. However, we are also going to use the p-value as the final comparison the with area of significance for our decision. Your p-value will be computed using either the z score or the t-statistic. The choice is yours.

2. The final step is to compute the p-value. You can do that by using the p-value calculator at the link below. You will calculate the p-value from “z” or “t” as a one tail test. The z/t score you will be using is 1.645.

3. https://goodcalculators.com/p-value-calculator/

For this test, we are comparing the p-value to the area of significance (0.05). Where does the p-value fall? What does that mean when we apply the decision rule? Now, explain what this tells you, i.e. what are the results of the test, and what would you recommend.