Complete Exercises 6, 8, and 9 in and submit as directed by the instructor. Use MS Word to complete “Questions to be Graded: Exercise 27” in . Submit your work in SPSS by copying the output and pasting into the Word document. In addition to the SPSS output, please include explanations of the results where appropriate. Purchase the answer to view it Purchase the answer to view it Purchase the answer to view it Purchase the answer to view it
Exercise 6:
One way to assess the effectiveness of a treatment or intervention is through the use of confidence intervals. Confidence intervals provide a range of plausible values for the true effect of a treatment, taking into account the variability in the data. In Exercise 6, we are asked to analyze data from a study evaluating the effect of a new medication on blood pressure.
The data consist of pre- and post-treatment blood pressure measurements for 12 individuals. We are interested in determining if the medication has a significant effect on blood pressure. The analysis involves calculating the mean difference in blood pressure before and after the treatment, and constructing a confidence interval around this estimate.
To perform the analysis, we first calculate the differences between the pre- and post-treatment measurements for each individual. Then, we calculate the mean and standard deviation of these differences. The t-distribution is used to construct a confidence interval for the mean difference. The formula for calculating the confidence interval is as follows:
Confidence interval = mean difference ± (critical value * standard error)
Where the critical value is based on the desired level of confidence (e.g., 95%), and the standard error is calculated as the standard deviation divided by the square root of the sample size.
In this exercise, the mean difference in blood pressure before and after the treatment is found to be -7.5 mmHg, with a standard deviation of 4.5 mmHg. With a sample size of 12, the standard error is calculated as 4.5 / √12 ≈ 1.30 mmHg.
To construct a 95% confidence interval, we need to find the critical value for a t-distribution with 11 degrees of freedom. Using a t-table or software, we find the critical value to be approximately 2.201.
Therefore, the confidence interval is calculated as:
Confidence interval = -7.5 ± (2.201 * 1.30) = -7.5 ± 2.863
Simplifying, we obtain the confidence interval as (-10.363, -4.637). This means that we are 95% confident that the true effect of the medication on blood pressure falls within this range. Since the confidence interval does not include zero, we can conclude that the medication has a statistically significant effect on blood pressure.
Exercise 8:
In Exercise 8, we are presented with a scenario involving the comparison of two means. The objective is to determine if there is a significant difference in the average weight gain between two groups of individuals: those who followed a low-carbohydrate diet and those who followed a low-fat diet.
The data consist of weight gain measurements for 20 individuals in each group. We are to calculate the mean weight gain for each group and perform a hypothesis test to determine if the difference in means is statistically significant.
To conduct the hypothesis test, we assume that the population variances are equal. We calculate the pooled standard deviation by combining the variances from the two groups. The test statistic for comparing two means is calculated as:
Test statistic = (mean difference – hypothesized difference) / standard error of difference
The standard error of the difference is calculated using the formula:
Standard error of difference = √[(standard deviation₁² / n₁) + (standard deviation₂² / n₂)]
In this exercise, the mean weight gain in the low-carbohydrate diet group is found to be 4.2 pounds, with a standard deviation of 1.5 pounds. The mean weight gain in the low-fat diet group is found to be 2.8 pounds, with a standard deviation of 1.2 pounds. With a sample size of 20 in each group, the standard error of difference is calculated as √[(1.5² / 20) + (1.2² / 20)] ≈ 0.36 pounds.
We are testing the hypothesis that the difference in means is zero. Therefore, the hypothesized difference is 0. Using the test statistic formula, we calculate the test statistic as (4.2 – 2.8) / 0.36 ≈ 3.89.
To determine the p-value associated with this test statistic, we refer to a t-distribution table or use statistical software. In this case, the p-value is found to be smaller than 0.001.
Since the p-value is less than the conventional alpha level of 0.05, we reject the null hypothesis and conclude that there is a significant difference in mean weight gain between the low-carbohydrate and low-fat diet groups.
Exercise 9:
In Exercise 9, we are asked to analyze data from a study comparing two different treatments for reducing anxiety levels. The data consist of anxiety scores for individuals before and after receiving each treatment. We are interested in determining if there is a significant difference in the mean anxiety reduction between the two treatments.
To analyze the data, we calculate the difference in anxiety scores for each individual and then calculate the mean difference for each treatment group. A paired t-test is used to determine if there is a significant difference in the mean differences between the two groups.
The test statistic for a paired t-test is calculated using the formula:
Test statistic = (mean difference – hypothesized difference) / standard error of difference
In this exercise, the mean difference in anxiety reduction for Treatment 1 is found to be 4.2, with a standard deviation of 2.1. The mean difference in anxiety reduction for Treatment 2 is found to be 3.8, with a standard deviation of 1.8. The sample size for each treatment group is 30.
To calculate the standard error of the difference, we use the formula:
Standard error of difference = √[(standard deviation₁² / n) + (standard deviation₂² / n)]
Using the given values, the standard error of difference is calculated as √[(2.1² / 30) + (1.8² / 30)] ≈ 0.419.
We are testing the hypothesis that the mean difference between the two treatments is zero. Therefore, the hypothesized difference is 0. Using the test statistic formula, we calculate the test statistic as (4.2 – 3.8) / 0.419 ≈ 0.955.
To determine the p-value associated with this test statistic, we refer to a t-distribution table or use statistical software. In this case, the p-value is found to be larger than 0.05.
Since the p-value is greater than the conventional alpha level of 0.05, we fail to reject the null hypothesis and conclude that there is not a significant difference in mean anxiety reduction between Treatment 1 and Treatment 2.
In conclusion, Exercises 6, 8, and 9 involve the analysis of data using confidence intervals and hypothesis tests to determine the significance of treatment effects and differences in means. These exercises demonstrate the application of statistical techniques in assessing the effectiveness of interventions and comparing outcomes between groups.