Question

Part 1.

The insulin pump is a device that delivers insulin to a diabetic patient at regular intervals. It presumably regulates insulin better than standard injections. However, data to establish this point are few, especially in children.

The following study was set up to assess the effect of use of the insulin pump on HbgA1c, which is a long-term marker of compliance with insulin protocols. In general, a normal range for HgbA1c is <7%.

Data were collected on 256 diabetic patients for 1 year before and after using the insulin pump. A subset of the data for 10 diabetic patients is given in the table below.

ID	HgbA1c 1 year before (%)	HgbA1c 1 year after (%)
1	6.7	7.0
2	7.4	7.4
3	9.2	8.6
4	9.6	8.1
5	7.4	6.8
6	8.1	7.0
7	10.8	8.5
8	7.1	7.7
9	7.9	9.7
10	10.8	7.7

Test the hypothesis that the measured HgbA1c 1 year after the insulin pump will be different from HgbA1c 1 year before the insulin pump. Set your p-value to indicate statistical significance at <0.05.

Type up a report that has the following information:

1. Write out the null and alternative hypothesis
2. Identify the types of variables that are being used – Nominal, ordinal, interval, or ratio?
3. Report the mean and standard deviation for each group
4. Display the data for each group using a histogram with the normal curve included (get the graph from SPSS and insert it into this document)
   1. List the conditions that should be met for conducting a hypothesis test of a mean
   2. Determine whether the data are appropriate for conducting a hypothesis test of a mean and indicate how you came to this conclusion.
5. Choose and identify the statistical test you will use to answer the hypothesis question assuming the conditions for conducting a hypothesis test of a mean are met
6. Indicate the alpha level, the df and whether it is one-tailed or two-tailed
7. Report t-value, p-value, and 95% confidence interval
8. Make a decision regarding the hypothesis
9. State your conclusion regarding the results of the study

Part II.

A clinical trial is conducted at the gynecology unit of a major hospital to determine the effectiveness of drug A in preventing premature birth, as indicated by infant birthweight. In the trial, 30 pregnant women are to be studied, 15 in a treatment group to receive drug A and 15 in a control group to receive a placebo. The patients are to take a fixed dose of each drug on a one-time-only basis between the 24^th and 28^th weeks of pregnancy. The patients are assigned to groups based on computer-generated random numbers, where for every two patients eligible for the study, one is assigned randomly to the treatment group and the other to the control group.

Table 1. Birthweights in a clinical trial to test a drug for preventing low-birthweight deliveries
Participant ID	Treatment Group	Birthweight
1	Drug A	6.9
2	Drug A	7.6
3	Drug A	7.3
4	Drug A	7.6
5	Drug A	6.8
6	Drug A	7.2
7	Drug A	8.0
8	Drug A	5.5
9	Drug A	5.8
10	Drug A	7.3
11	Drug A	8.2
12	Drug A	6.9
13	Drug A	6.8
14	Drug A	5.7
15	Drug A	8.6
16	Placebo	6.4
17	Placebo	6.7
18	Placebo	5.4
19	Placebo	8.2
20	Placebo	5.3
21	Placebo	6.6
22	Placebo	5.8
23	Placebo	5.7
24	Placebo	6.2
25	Placebo	7.1
26	Placebo	7.0
27	Placebo	6.9
28	Placebo	5.6
29	Placebo	4.2
30	Placebo	6.8

Test the hypothesis that drug A produces significantly different birthweight as compared to a placebo. Set you p-value to indicate statistical significance at p<0.05.

Type up a report that has the following information:

1. Write out the null and alternative hypothesis
2. Identify the types of variables that are being used – Nominal, ordinal, interval, or ratio?
3. Report the mean and standard deviation for each group
4. Display the data for each group using a histogram with the normal curve included (get the graph from SPSS and insert it into this document)
   1. List the conditions that should be met for conducting a hypothesis test of a mean
   2. Determine whether the data are appropriate for conducting a hypothesis test of a mean and indicate how you came to this conclusion.
5. Choose and identify the statistical test you will use to answer the hypothesis question assuming the conditions for conducting a hypothesis test of a mean are met
6. Indicate the alpha level, the df and whether it is one-tailed or two-tailed
7. Report t-value, p-value, and 95% confidence interval
8. Make a decision regarding the hypothesis
9. State your conclusion regarding the results of the study

Answer 1

The insulin pump is a device that delivers insulin to a diabetic patient at regular intervals.

	Sample 1	Sample 2	Difference = Sample 1 - Sample 2
	6.7	7	-0.3
	7.4	7.4	0
	9.2	8.6	0.6
	9.6	8.1	1.5
	7.4	6.8	0.6
	8.1	7	1.1
	10.8	8.5	2.3
	7.1	7.7	-0.6
	7.9	9.7	-1.8
	10.8	7.7	3.1
Average	8.5	7.85	0.65
St. Dev.	1.506	0.898	1.435
n	10	10	10

From the sample data, it is found that the corresponding sample means are:

Xˉ1=8.5 ; Xˉ2=7.85

Also, the provided sample standard deviations are:

s1=1.506 ; s2=0.898

and the sample size is n = 10. For the score differences we have

Dˉ=0.65 ; s_D =1.435

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μD = 0

Ha: μD ≠ 0

This corresponds to a two-tailed test, for which a t-test for two paired samples be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the degrees of freedom are df = 9.

Hence, it is found that the critical value for this two-tailed test is tc=2.262, for α=0.05 and df = 9

The rejection region for this two-tailed test is,

R={t:∣t∣>2.262}.

(3) Test Statistics

The t-statistic is computed as shown in the following formula:

(4) Decision about the null hypothesis

Since it is observed that ∣t∣=1.432≤tc​=2.262, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.186, and since p=0.186≥0.05, it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μ1 is different than μ2, at the 0.05 significance level.

Confidence Interval

The 95% confidence interval is −0.377<μD​<1.677.

Graphically,

Part - II

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ1 = μ2

Ha: μ1 ≠ μ2

This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the degrees of freedom are df = 27.88df=27.88. In fact, the degrees of freedom are computed as follows, assuming that the population variances are unequal:

Hence, it is found that the critical value for this two-tailed test is tc=2.049, for α=0.05 and df = 27.88.

The rejection region for this two-tailed test is

R={t:∣t∣>2.049}.

(3) Test Statistics

Since it is assumed that the population variances are unequal, the t-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that ∣t∣=2.414>tc​=2.049, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0.0226, and since p=0.0226<0.05, it is concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population mean μ1 is different than μ2, at the 0.05 significance level.

Confidence Interval

The 95% confidence interval is 0.124<μ1​−μ2​<1.516.