Question

Question Number 3: A new drug is proposed to lower total cholesterol and a study is designed to evaluate the efficacy of the drug in lowering cholesterol. Fifteen patients agree to participate in the study and each is asked to take the new drug for 6 weeks. However, before starting the treatment, each patient's total cholesterol level is measured. The initial measurement is a pre-treatment or baseline value. After taking the drug for 6 weeks, each patient's total cholesterol level is measured again and the data are shown below. Subject Identification Number   Baseline   6 Weeks 1   215   205 2   190   156 3   230   190 4   220   180 5   214   201 6   240   227 7   210   197 8   193   173 9   210   204 10   230   217 11   180   142 12   260   262 13   210   207 14   190   184 15   200   193 Is there statistical evidence of a reduction in mean total cholesterol in patients after using the new medication for 6 weeks?

Answer 1

First of all note that the data has been collected on the same set of people before and after treatment with new drug. As the data has been collected using repeated samples, so the data is paired and thus we would use a paired sample t-test.

As we want to test if the drug has reduced the mean cholestrol in patients, so we would want the mean of differences for population to be positive such that the data before treatment are higher than post treatment.

Define
11
and
112
as the population mean cholestrol level before and after administration of drug into the body respectively. And let
D
be the difference -

HD 11 H2

So, the hypothesis would be lower tailed written as -

Null Hypothesis - H₀:

HD=0

Alternate Hypothesis - H₁:

07 01

Now, create a separate column for difference between the values of 'Baseline' and '6 Weeks After' as shown -

Identification Number	Baseline (X1i)	6 Weeks after (X2i)	Difference (di)
1	215	205	10
2	190	156	34
3	230	190	40
4	220	180	40
5	214	201	13
6	240	227	13
7	210	197	13
8	193	173	20
9	210	204	6
10	230	217	13
11	180	142	38
12	260	262	-2
13	210	207	3
14	190	184	6
15	200	193	7

Now, you can either use excel to directly get the test result or you can do it manually.

1) Using Excel

Use excel's 'Data Analysis' add-on under the 'Data' tab to access the 't-Test: Paired Two Sample for Means' as shown -

File Home Insert Draw Page Layout Formulas Data Review View Help Power Pivot Share From Text/CSV Queries & Connections I UN Recent Sources [9 Existing Connections A1 MA 如密 Properties Clear Reapply Advanced Data Analysis ? Solver From Web Get Data From Table/Range Group Ungroup Subtotal Stocks Geography Sort Refresh All Filter Text to What-If Forecast Analysis Sheet Edit Links Columns EO Get & Transform Data Queries & Connections Data Types Sort & Filter Data Tools Forecast Outline Analyze J1 fx חד E J K L M N O Р Q R S T U V w x 1 1 2 3 Data Analysis ? Х 4 OK Cancel 5 6 7 8 9 G H Identification Number Baseline 6 Weeks after Difference 215 205 10 2 190 156 34 3 230 190 40 4 220 180 40 5 214 201 13 6 240 227 13 7 210 197 13 8 193 173 20 9 210 204 6 10 230 217 13 11 180 142 38 12 260 262 -2 Analysis Tools F-Test Two-Sample for Variances Fourier Analysis Histogram Moving Average Random Number Generation Rank and Percentile Regression Sampling t-Test: Paired Two Sample for Means t-Test: Two-Sample Assuming Equal Variances Help 10 11 12 13 14 13 210 3 207 184 15 14 190 6 15 200 193 7 16 17

Then in the next prompt box, enter the variable ranges as the 'Baseline' and '6 Weeks Later' data. As the hypothesized mean difference is '0', so put the same in prompt. Also make sure to mark the 'Labels' option as we have label in our first row. Your input should look similar to this -

J1 X E S t-Test: Paired Two Sample for Means ? X 1 2 ܢܙ Input Variable 1 Range: OK 3 $G$1:$G$16 1 4 Variable 2 Range: Cancel $H$1:$H$16 1 5 Help 6 Hypothesized Mean Difference: ol 7 Labels 8 F G H Identification Number Baseline 6 Weeks after Difference 215 205 10 2 190 156 34 3 230 190 40 4 220 180 40 5 214 201 13 6 240 227 13 7 210 197 13 8 193 173 20 9 210 204 6 10 230 217 13 11 180 142 38 12 260 262 -2 13 210 207 3 14 190 184 6 15 200 193 7 Alpha: 0.05 9 10 Output options O Qutput Range: 11 1 12 O New Worksheet Ply: 13 O New Workbook 14 15 16 17 18 19

Note that you can set level of significance as per your needs. By default its chosen as 0.05.

This should give you following result -

t-Test: Paired Two Sample for Means
	Baseline	6 Weeks after
Mean	212.8	195.8666667
Variance	450.8857143	824.2666667
Observations	15	15
Pearson Correlation	0.881282389
Hypothesized Mean Difference	0
df	14
t Stat	4.630004243
P(T<=t) one-tail	0.000194797
t Critical one-tail	1.761310136
P(T<=t) two-tail	0.000389595
t Critical two-tail	2.144786688

As we had a one-tailed test, so note that the p-value for a one tailed test = 0.000194797 $\approx$ 0.0002

As the p-value is less than the significance level, so we reject the null hypothesis and conclude that there is enough evidence to support the claim that the cholestrol level has decreased 6 weeks after administration of drug in the body.

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Solving it manually -

Calculate the sample mean differences and standard deviation of differences using formula -

$\overline{X_{D}} = \frac{\sum d_{i}}{n}$

$s_{D} = \sqrt{\frac{\sum (d_{i} - \overline{X_{D}})}{n-1}}$

You can calculate the values as -

Identification Number	Baseline (X1i)	6 Weeks after (X2i)	Difference (di)	$d_{i} - \overline{X_{D}}$	$(d_{i} - \overline{X_{D}})^2$
1	215	205	10	-6.93333	48.07111
2	190	156	34	17.06667	291.2711
3	230	190	40	23.06667	532.0711
4	220	180	40	23.06667	532.0711
5	214	201	13	-3.93333	15.47111
6	240	227	13	-3.93333	15.47111
7	210	197	13	-3.93333	15.47111
8	193	173	20	3.066667	9.404444
9	210	204	6	-10.9333	119.5378
10	230	217	13	-3.93333	15.47111
11	180	142	38	21.06667	443.8044
12	260	262	-2	-18.9333	358.4711
13	210	207	3	-13.9333	194.1378
14	190	184	6	-10.9333	119.5378
15	200	193	7	-9.93333	98.67111
Sum			254		2808.933

$\overline{X_{D}} = \frac{\sum d_{i}}{n} = \frac{254}{15} = 16.9333$

$s_{D} = \sqrt{\frac{2808.933}{14}} = 14.164668$

Then the test statistic is -

$t = \frac{\overline{X_{D}} - \mu_{D}}{\frac{s_{D}}{\sqrt{n}}}$

So, we get -  

$t = \frac{16.9333 - 0}{\frac{14.16468}{\sqrt{15}}} = 4.630$

The degree of freedom of test = n-1 = 15 - 1 = 14

Significance level = $\alpha$ = 0.05

So, critical value of test statistic = t_0.05,14 = 1.761

As calculated value of test statistic = 4.63 > critical value of test statistic = 1.761, so we reject the null hypothesis and conclude that there is enough evidence in the data to support the claim that the mean cholestrol of population decreases after 6 weeks of administration of drug into body.

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Please ask if you have any doubt(s) in comment section.

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