Healthcare SLP 4 part 2

Running head: STATISTICS 1

STATISTICS 3

Statistics

Student’s Name

Institutional Affiliation

Question 1

The first step is to the hypotheses of the study. After formulating the hypothesis and determining the significance. The next step is to determine the test statistics. In the study, a chi-squared test will be used to test the hypothesis (Cook, Netuveli, & Sheikh, 2004). The expected values are calculated from the observed values to determine how the population would be if there was no relationship if cholesterol treatment is ineffective. The expected values will be used to test the hypothesis using p-value from the chi-squared statistics. A low p-value means that cholesterol treatment is effective while a high p-value shows that the treatment is ineffective and there is no difference in the level of cholesterol in the population (Cook, Netuveli, & Sheikh, 2004).

Question 2

The null hypothesis states that there is no relationship between the cholesterol treatment and the increase/decrease in the cholesterol level (Cook, Netuveli, & Sheikh, 2004). The test will be conducted to disapprove the null hypothesis. The alternative hypothesis states that the cholesterol treatment is effective. The selected significance level 0.05 or 5%, a p-value smaller than 0.05 means cholesterol treatment is effective. Cook, Netuveli, & Sheikh (2004), “the null hypothesis should be rejected if the p-value is smaller than the significance level. (Cook, Netuveli, & Sheikh, 2004). The null hypothesis (H0) is states as: treatment has no effect on the cholesterol level is not effective and the alternative hypothesis (H1) is states as: treatment for cholesterol is effective

Question 3

The study will use the chi-square test to determine if the cholesterol treatment is effective. According to Cook, Netuveli, & Sheikh (2004), the chi test formula = , where c is the degree of freedom, O represents the observed participants and E is the expected participants in the sample(Cook, Netuveli, & Sheikh, 2004). The chi test is used to determine the relationship between the participants who received treatment and who didn’t (Cook, Netuveli, & Sheikh, 2004). A chi test provides a single value to show the difference between the observed and the number of counts that would be expected if there was no relationship between treatment and the level of cholesterol among the participants.

Question 4

Table 1- observed data of 114 participants

	Cholesterol Decreased	No Cholesterol Decrease	Total
Treatment	38	18	56
No treatment	30	28	58
Total	68	46	144

The observed data is used to calculate the expected values (Cook, Netuveli, & Sheikh, 2004). Each expected value is calculated using the following formula; [(column value x Row total)/ column total/grand total]

Table 2-Expected

	Cholesterol Decreased(A)	No Cholesterol Decrease (B)	Total (C)
Treatment (D)	=33.40	=18	56
No treatment (E)	=34.60	=28	58
Total (F)	68	46	144

The difference between the observed and the expected is (38-33.40) =4.6.

Chi-squared test=

Chi test=+++=3.17

The formula for calculating the degrees of freedom (rows-1) x (colunms-1) =1.

From the Chi statistics the p-value=0.0750 (0.75%)

Question 5

A small p-value means that there is a large difference between the individuals in the sample and the expected participants (Norman, & Streiner, 2014). Therefore, the results of the sample are different from the expected results if cholesterol treatment is not effective (Cook, Netuveli, & Sheikh, 2004). A small p-value of 0.0750 means that chance of cholesterol treatment being ineffective is 0.75%. This provides strong evidence that treatment for cholesterol is effective (Norman, & Streiner, 2014). Therefore, there is enough evidence to reject the null hypothesis. A large p-value would mean that the cholesterol treatment is ineffective.

References

Cook A., Netuveli, G., & Sheikh, A. (2004). Basic skills in statistics: A guide for healthcare professionals. London, GBR: Class Publishing. eISBN: 9781859591291.

Norman, G. R., & Streiner, D. L. (2014).Biostatistics: The Bare Essentials [4th ed., e-Book]. Shelton, Connecticut: PMPH-USA, Ltd. eISBN-13: 978-1-60795-279-4

NEW HOMEWORK ASSIGNMENT

Module 4 - SLP

DRAWING INFERENCES ABOUT POPULATION MEANS AND PROPORTIONS

Having developed the null and alternative hypotheses in the previous module, write a (2-3 pages) paper in which you:

1. Identify a test statistic to help you assess the evidence against the null hypothesis you developed in the previous module.

2. Explain why you have chosen the specific test statistic. Include in your discussion description of the test statistic.

3. Summarize your findings by creating a summary graph in which you display your data.

4. Discuss the total number of measurements (sample size), the possibility for measurement error, and whether it is large enough to paint an accurate picture.