Hypothesis Testing Problem Worksheet

Introduction:

In the field of statistics, hypothesis testing is a powerful tool used to make decisions or draw conclusions about a population based on a sample. It involves formulating two mutually exclusive hypotheses, the null hypothesis (H0) and the alternative hypothesis (Ha), and testing them using statistical methods. The objective is to determine whether there is sufficient evidence to reject the null hypothesis and support the alternative hypothesis.

Problem 1:

A researcher is interested in studying the effect of a new medication on blood pressure. The null hypothesis is that the medication has no effect, while the alternative hypothesis is that the medication reduces blood pressure. The researcher collects data from a random sample of 100 participants and calculates their blood pressure before and after taking the medication. The mean difference in blood pressure is 5 mmHg with a standard deviation of 3 mmHg.

To test the hypotheses, the researcher uses a two-tailed independent samples t-test with a significance level of α = 0.05. The t-value is calculated as:

t = (mean difference – hypothesized mean difference) / (standard deviation / √n)

where n is the sample size. In this case, the hypothesized mean difference is 0, as the null hypothesis states that the medication has no effect.

The critical values for the t-test are determined based on the degrees of freedom, which is equal to the sum of the individual degrees of freedom for the two samples. For an independent samples t-test, the degrees of freedom can be calculated as:

df = (n1 – 1) + (n2 – 1)

where n1 and n2 are the sample sizes of the two groups being compared.

The decision to reject or fail to reject the null hypothesis is based on comparing the calculated t-value with the critical t-values. If the calculated t-value falls within the acceptance region (between the critical t-values), the null hypothesis is not rejected. If the calculated t-value falls in the rejection region (beyond the critical t-values), the null hypothesis is rejected in favor of the alternative hypothesis.

In this problem, the researcher calculates the t-value to be (5 – 0) / (3 / √100) = 5 / (3 / 10) = 16.67. The degrees of freedom are 100 – 1 = 99, as there are 100 participants in the sample.

The critical t-values for a two-tailed test with α = 0.05 and 99 degrees of freedom can be obtained from a t-table or using statistical software. In this case, the critical t-values are approximately -2.62 and 2.62.

Since the calculated t-value (16.67) falls in the rejection region (outside the range of -2.62 to 2.62), the null hypothesis is rejected. There is sufficient evidence to conclude that the new medication reduces blood pressure.

Problem 2:

A manufacturing company is interested in comparing the mean breaking strength of two different materials used in production. The null hypothesis is that the mean breaking strength is the same for both materials, while the alternative hypothesis is that there is a difference in mean breaking strength.

The company collects data from two random samples of 50 specimens each and calculates the sample means and standard deviations. The mean breaking strength for material A is 1000 N with a standard deviation of 50 N, while the mean breaking strength for material B is 1100 N with a standard deviation of 60 N.

To test the hypotheses, the company uses a two-tailed independent samples t-test with a significance level of α = 0.01. The t-value is calculated using the formula mentioned earlier.

The degrees of freedom for this test can be calculated as (50-1) + (50-1) = 98.

The critical t-values for a two-tailed test with α = 0.01 and 98 degrees of freedom are approximately -2.62 and 2.62.

The calculated t-value is (1000 – 1100) / (√[(50^2 / 50) + (60^2 / 50)]) = -100 / (√[2500/50 + 3600/50]) = -100 / (√50 + 72) ≈ -5.67.

Since the calculated t-value (-5.67) falls in the rejection region, the null hypothesis is rejected. There is sufficient evidence to conclude that there is a difference in mean breaking strength between the two materials.

Conclusion:

Hypothesis testing is an essential tool in statistics for making informed decisions based on sample data. By formulating null and alternative hypotheses and conducting appropriate statistical tests, researchers can determine whether there is sufficient evidence to support or reject a particular hypothesis. This process allows for confident conclusions to be drawn about populations based on limited sample data.

When conducting hypothesis testing, it is important to choose an appropriate significance level and use the correct statistical test based on the nature of the data and research question. Additionally, understanding the concept of critical values and how to interpret test statistics such as t-values is vital for accurate hypothesis testing.

Overall, hypothesis testing provides a rigorous methodology for making statistical inferences and is widely used in various fields of research and decision-making.