Drawing Inferences

**QUESTION**

A researcher conducts a test on the effectiveness of a cholesterol treatment on 114 total subjects. Assuming the tails of distributions are normal distribution, is there evidence that the treatment is effective?

Cholesterol Decreased

No Cholesterol Decrease

Total

Treatment

38

18

56

No treatment

30

28

58

Total

68

46

114

Submit a (2-3 pages) paper before the end of the module that:

Briefly describe the procedure for testing hypotheses.

Formulate the null and alternative hypotheses.

Determine the test statistic that you will use.

Calculate the p-value. Show your work.

Discuss whether there is enough evidence to reject the null hypothesis.

Drawing Inferences

**ANSWER**

Drawing Inferences

Name

Course

Professor’s name

Institution

Date

Cholesterol Decreased No Cholesterol Decrease Total

Treatment 38 18 56

No Treatment 30 28 58

Total 68 46 114

Since the variables involved in this test are categorical, I will use the chi-square test of Independence to test hypotheses. Chi-square test in instances where the data contains categorical variables from a sample or population, hence there is a requirement to check whether there exists a significant relationship between these two variables (Qian, 2009).

This test procedure is applicable only when these outlined conditions are satisfied: The method used to sample was simple random sampling. The variables involved in the research are all categorical. If the sample data is displayed on a contingency table, we expect the frequency count for every individual cell in the table to be less than 5.

This method is made up of four simple steps: (1) stating the hypotheses, (2) formulating an analysis plan, (3) analyzing sample data, and finally (4) interpreting the results. Under stating your hypotheses, we apply the concept that a variable A has m levels, and a variable B has n levels. The null hypothesis means that knowing the level of the variable A does not aid you in predicting the variable B level. That is, the variables are not related, or rather they are independent. The null hypothesis, H0, states that variable A and variable B are independent. The alternative hypothesis, H1, states that variable A and Variable B are related to each other. In other words, this alternative hypothesis means that being aware of variable A’s level can aid you in predicting variable B’s level (Wilcox, 2017).

In this instance, the following statement represents the null hypothesis (H0) for this test;

H0: Cholesterol treatment and Cholesterol levels are not dependent on each other

The following statement represents the alternative hypothesis (H1) for this test;

H1: Cholesterol treatment and cholesterol levels are dependent

The test statistic for the Chi-square test of independence is calculated using the following formula; ꭓ2 = ∑ [(O – E)2 /E]. Here, O represents the observed values, and E represents the expected values. To arrive at the chi-squared statistic value, note the difference between observed values (O) and expected values (E) in pairs, square their differences, and divide the squared differences by their respective expected values. Redo this procedure for all the cells in the contingency table, and lastly, sum the values. The result value is the χ2 (Qian, 2009).

We first need to compute the expected values for each table cell. The formula to use here is; Expected value for each table cell = (Row total * Column total) / Sample Size (Qian, 2009). To calculate our expected frequency for the Treatment/Cholesterol Decreased cell in this dataset, we shall follow these steps: first, identify the row total for Treatment (56), secondly identify the column total for Cholesterol Decreased (68), and lastly multiply the two values, and then divide the multiplication result by the total sample size (114). That is, (56 * 68) / 114 = 33.404. After obtaining all the expected values, we will then use these values to compute the chi-square test statistic using the previous paragraph formula.

The following table represents the results for the observed values, the expected values, and the squared difference between the observed and the expected values divided by the expected values;

Cholesterol Level Cholesterol Treatment Observed (O) Expected (E) [(O – E)2 /E

Cholesterol Decreased Treatment 38 (56*68)/114 = 33.404 0.632

Cholesterol Decreased No Treatment 30 (58*68)/114 = 34.596 0.611

No Cholesterol Decrease Treatment 18 (56*46)/114 = 22.596 0.935

No Cholesterol Decrease No Treatment 28 (58*46)/114 = 23.404 0.903

Therefore, ꭓ2 = ∑ [(O – E)2 /E] = 0.632+0.611+0.935+0.903 = 3.081

After obtaining our test statistic, the next step is to compute the p-value. Therefore, we need to calculate the degrees of freedom to achieve this. For a table consisting of m rows and n columns, the formula for computing degrees of freedom for the chi-square test is (m-1) (n-1). In this case, we only have two rows and two columns: (2-1) * (2-1) = 1 degree of freedom. To obtain the p-value, we are going to use the X2 distribution table. Having 1 degree of freedom and assuming a significance level of α = 0.05, we get that our p-value = 0.0792

Because the obtained p-value is greater than ‘either 0.05 level of significance, there is not enough evidence to reject the null hypothesis. Therefore, we fail to reject the null hypothesis (Hartshorn, 2017). The conclusion drawn from these test results is that the Treatment of Cholesterol and Cholesterol levels are not dependent on each other. This means that a change in the Treatment of Cholesterol will not affect the levels of Cholesterol.

References

Hartshorn, S. (2017). Hypothesis testing: A visual introduction to statistical significance.

Qian, H. (2009). On data-driven chi-square statistics.

Wilcox, R. R. (2017). Introduction to robust estimation and hypothesis testing.