Posted: February 12th, 2015

Biostatistics Hw

Paper, Order, or Assignment Requirements

 

 

Homework on 2 sample tests.

 

1. Create SPSS data files with the data presented in class (n of 5, paired samples) and n of 10, independent samples.
2. Confirm the test results presented
3. contrast the results of the two tests of the hypothesis—was one design better than the other?

 

You continue work on the treatment of severe xerostomia (low salivary flow rates) in head/neck cancer patients treated receiving radiation. In addition to the data you already collected on the pre-post efficacy of the new lozenge, Drool, new data are now available showing salivary flow and subjective ratings of dryness before and after treatment on the standard of care, a product called WetMouth. As before, abnormal flow is defined as < 1 mL/min, and subjective reports of xerostomia were obtained from a 10-point VAS anchored by ‘worst dry mouth imaginable (0)’ and ‘no problem with dry mouth (9)’. Using the ‘dry mouth 2-sample’ data file to answer the following:

 

2. Determine the xerostomia rate at baseline in the control sample. Compare it to the Drool sample. Do you need to do a statistical test? Why?
3. Determine the median rating of xerostomia at follow-up in each treatment arm. Is the final rating of xerostomia different in the head/neck cancer patients randomized to receive Drool than the patients randomized to receive WetMouth? Conduct a traditional test of the hypothesis that the median in the Drool sample is less than that in the WetMouth sample (α= .05). Include the following in your summary of the analysis:
4. The median, IQR, M and SD ratings at follow-up.
5. The ES describing the difference in means.
6. The results of your statistical test.
7. A statement about the equality of error variances.
8. The power of your test.
9. The sample size that would be needed to achieve power of 80%.
10. relevant output from SPSS and/or G*Power.

 

4. Repeat Q3, but use the flow data as the outcome measure.

 

5. While not true in this instance, unequal baseline scores in the control and treatment groups make the comparison of the post-treatment scores difficult. One simplifying mechanism is to compare the groups on their difference scores.
6. Use SPSS to compute new variables showing the difference between pre and post scores for the measure of salivary flow.
7. Set up a traditional hypothesis test of the difference between the treatment groups means. Choose the significance level and justify it.
8. Write up the results of the analysis, addressing a-g, as in Q1.

 

 

Biostatistic homework 5

1a. xerostomia rate =72%

 

H₀ : Xerostomia rate in this sample is equal to national rate (0.75) in head and neck cancer patients

H_A : Xerostomia rate in this sample is less than national rate (0.75) in head and neck cancer patients

α=0.05

 

Assumptions: data is normally distributed; observations are independent

 

Test statistic : Student’s t-test

From the output, t=-32733, p=0.7463>0.5, we do not reject the null hypothesis . We do not have enough evidence from these data to reject the null and accept the alternative hypothesis that the xerostomia rate in this sample is less than that of the national rate (0.75) at 5% level of significance. The observed differences between the sample xerostomia rate (0.72) and the national xerostomia rate (0.75) is due to chance.

 

 

1b. xerostomia rate =72%

Assumptions: data is normally distributed; observations are independent

 

H₀ : Xerostomia rate in this sample is equal to the national rate (0.55) in head and neck cancer patients

H_A : Xerostomia rate in this sample is less than national rate (0.55) in head and neck cancer patients

α=0.01

 

Test statistic: Student’s t-test

 

From the output, t=1.8549, p=0.0759 <0.10, we reject the null hypothesis. We have enough evidence from these data to reject the null and accept the alternative hypothesis that the xerostomia rate in this sample is less than that of the national rate (0.55) at 1% level of significance.

 

2a. Median baseline xerostemia =2

 

H₀ : Median Xerostomia rating in this sample is equal to the national rate (3) in head and neck cancer patients

H_A : Median Xerostomia rating in this sample is less than national rate (3) in head and neck cancer patients

α=0.05

Assumptions: data is normally distributed; observations are independent

 

Test statistic: Student’s t-test

 

 

From the output, t=-3.61158, p=0.0014 <0.05, we reject the null hypothesis. We have enough evidence from these data to reject the null and accept the alternative hypothesis that the median xerostomia rating in this sample is less than the population rating at 5% level of significance.

 

2b. M(SD) rating is 2±1.38 at baseline. Comparing this to the national median (3) It supports the median test.

 

3. Descriptive statistics for baseline rating

 

 

Descriptive statistics for follow up rating

 

 

The post treatment (follow up rating) rating have a higher mean (mean=2.6) than the baseline ratings (mean=2). The baseline ratings have a higher spread ( std=1.38, variance=1.92) than the follow-up(std=1.15, variance=1.33), which shows a larger spread of data for baseline ratings.

Both the baseline and post treatment rating are somewhat slightly positively skewed with baseline rating slightly higher (skewness = 0.614 and 0.529 respectively). Both the baseline and the post treatment are platykurtic (kurtosis=0.053 and -0.240 respectively. There are no outliers in both distributions. Both distributions are somewhat normally distributed.

3a.

H₀ : Drool does not increase subjective ratings of xerostomia in head and neck cancer patients

H_A : Drool increases subjective ratings of xerostomia in head and neck cancer patients

α=0.05

 

Assumptions: data is normally distributed; observations are independent , one variable is a continuous variable (ratings) and another is a categorical variable (baseline or post-treatment)

Test statistic: Student’s t-test: 2-sample ttest

 

 

From the output, pooled t=2.94 p=0.0074< 0.05, we reject the null hypothesis. We have enough evidence from these data to reject the null and accept the alternative hypothesis that Drool significantly increases subjective ratings of xerostomia at 5% level of significance.

 

 

1. Without assuming level of measurement

 

H₀ : Drool does not increase subjective ratings of xerostomia in head and neck cancer patients

H_A : Drool increases subjective ratings of xerostomia in head and neck cancer patients

α=0.05

 

Assumptions: data is normally distributed; observations are independent

 

Test statistic: Student’s t-test:

 

 

 

From the output, t=11.26 p<0.0001< 0.05, we reject the null hypothesis. We have enough evidence from these data to reject the null and accept the alternative hypothesis that Drool significantly increases subjective ratings of xerostomia at 5% level of significance.

 

Assuming interval measures will keep us from getting an exaggerated effect of Drool in increasing subject ratings of xerostomia. Therefore comparing with the baseline helps us establish the true effect of Drool.

 

4. Descriptive statistics of baseline flow

 

 

 

Descriptive statistics of follow up flow

 

 

Follow up saliva flow is higher than baseline flow (mean=1.2 ml/min compared to 0.82 ml/min). Baseline flow has a higher standard deviation (std=0.310) compared to follow up flow (std=0.204) which shows a higher spread of data for baseline flow.

The baseline flow is positively skewed (skewness=2.49) and leptokurtic (kurtosis=4.56) while the follow up flow is somewhat slightly positively skewed (skeweness=0.529) and is platykurtic (kurtosis=-0.24). There were no obvious outliers but the baseline flow data seems not to be normally distributed.

4a. Assuming data is normally distributed;

 

H₀ : Drool does not increase flow of saliva in head and neck cancer patients

H_A : Drool increases flow of saliva in head and neck cancer patients

α=0.05

 

 

Assumptions: data is normally distributed; observations are independent

 

 

From the output, t=29.39 p<0.0001< 0.05, we reject the null hypothesis. We have enough evidence from these data to reject the null and accept the alternative hypothesis that Drool significantly increases flow of saliva at 5% level of significance.

 

4b. Assuming the data is not normally distributed :

H₀ : Drool does not increase flow of saliva in head and neck cancer patients

H_A : Drool increases flow of saliva in head and neck cancer patients

α=0.05

 

 

From the output, t=12.96 p<0.0001< 0.05, we reject the null hypothesis. We have enough evidence from these data to reject the null and accept the alternative hypothesis that Drool significantly increases flow of saliva at 5% level of significance.

If the data is not normally distributed, the efffect of Drool on saliva flow is significanlty reduced as shown by the two t-values.