Introduction

For a hypothesis testing to be carried out below are some assumptions about the data set are made. They include:

• Equal interval or ratio data
• Normal distribution or closely so
• Homogeneity of the variance-similar variance in each group
• Samples randomly were drawn from the population

N/B: If any of the assumptions 1, 2, 3and 4 are not met then non-parametric tests should be used.

Questions:

1. An example of a hypothesis where a one-tailed hypothesis test would be used

On the basis of the random sample X₁, X₂, X₃,… X_n. One can test the hypothesis concerning the value of . The null and alternative hypothesis can be formulated as below.

Scenario1:       H_o:  =µ_o                                            H₁: >µ_o

Scenario 2:      H_o:  =µ_o                                            H₁: <µ_o

Where µo is a specified value

The critical region R is given by R: | -µ_o | ≥ c and in the two situations it can be illustrated as

Example

In a continuous production process, it was found that the mean measurement of some characteristics of product occasionally shifts due to slight changes in machine setting, while the variability is seldom affected. The periodic checkups are made to ensure that the mean production is not off the mark and  is stable. Suppose that in this production process the target value of  is µ_o = 50 and  is known to be 2.5. The sample measurement are; 43,51,50,41,53,52,47,54,51,45,38 and 47. The production manager will welcome any changes towards higher values however he will safeguard against decreasing values . Formulate the null hypothesis and test the same.

Solution:

As the target value of 50, the null hypothesis is H_o:  =50. Since the production manager wants to safeguard against the decreasing values of the mean alternative hypothesisis H₁: >µ_o. to test the null hypothesis we calculate the value of the normal variate;

=

Given; , n=12, and

= -3.24

For 5% level of significance the critical region;

The calculated value of Z = -3.24< -1.796, so the null hypothesis is rejected at 5% level of significance. This means that some action is needed because  is significantly less than 50

2. An example of a hypothesis where two-tailed hypothesis test would be used

Situation 1:      H_o:  =µ_o                                            H₁:  ≠µ_o

Example

3. If a researcher has set alpha at 0.05 for a two-tailed hypothesis test, what is the p-value required to reject the null hypothesis

In hypothesis testing, p-value is compared to the alpha-value to determine if the data is statistically significantly different from what the null hypothesis states. In the case, if the p-value by any chance would be less than or equal to the alpha value, that is p < 0.05, then the null hypothesis will be rejected and a conclusion stating that the results are statistically significant be made.

4. A researcher has set alpha at 0.05. When the researcher analyzes the data from the experiment using a software program, she obtains a p-value equal to 0.932. Based on this p-value, should the researcher reject the null hypothesis or fail to reject the null hypothesis? Please explain your answer

P value = 0.932           alpha value = 0.05

Alpha value states how extreme the data must be before we reject the null hypothesis while the p value states how extreme the data is. Therefore in the scenario, the p-value of 0.932 is greater than the alpha value (p> 0.05), then we fail to reject the null hypothesis

Conclusion: The results are statistically non-significant.

5. A researcher has set alpha at 0.01. When the researcher analyzes the data from the experiment using a software program, he obtains a p-value equal to 0.04. Based on this p-value, should the researcher reject the null hypothesis or fail to reject the null hypothesis? Please explain your answer

Alpha = 0.01         p-value = 0.04

Alpha value states how extreme the data must be before we reject the null hypothesis while the p value states how extreme the data is. Therefore, in the scenario, the p-value of 0.04 is greater than the alpha value (p> 0.01), then we fail to reject the null hypothesis

Conclusion: The results are statistically non-significant.

6. A researcher is interested in whether music played during an exam will improve exam performance. Students in one class listen to music during an exam and students in another class take the exam in silence. The researcher set alpha at 0.05. Test scores for both classes are compared using a statistical software program. The mean test score for the class that listened to music during the exam is 95, while the mean test score for the class that took the exam in silence is 82. The obtained p-value from the independent group’s t-test is 0.02.
7. State the null and alternative hypothesis

H₀: There no difference between the performances of the student who took the exam when there were music and student who took the exam in silence

1. Determine if this is a directional or non-directional test. Please explain your answer.

Directional test – There is a decrease and an increase in performance away from the mean. There is a decrease between the student who took the exam in silence and those who had music on.A decrease of 13 marks.

1. Establish the conclusion of this study based on the p-value and the means provided.

The p-value is less than the alpha value, so we reject the null hypothesis

Conclusion:  The results are statistically significant.

1. Describe the Type I and type II error for this study.

Type I error, – will occur when the null hypothesis is true but it’s rejected.

Reject the null hypothesis that there is no difference in the performance and yet it is true

The student performed when the music was on

Type II error – will occur when the null hypothesis is accepted, and it’s false.

Accept that there is a differencein performanceand its false

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