Question

the cholesterol level of patients who had heart attacks was measured two days after the heart attack and then again four days after the heart attack. the researchers want to see if the cholesterol level of patients who have heart attacks reduces as the time since their heart attack increases. the data is in the table. calculate a 99% confidence interval for the mean difference in cholesterol levels from day two to day four. round to two decimal places.

cholesterol levels (in mg/dl) of heart attack patients

cholesterol level day 2	cholesterol level day 4
210	183
300	260
286	249
288	264
231	191
281	251
168	151
315	323
316	327
338	347
259	262
236	221
211	228
234	257
235	212
277	273
269	260
288	295
245	235
362	326
258	283
251	211
228	249
314	294
277	238
311	314
165	173

with % confidence, it can be said that the true mean difference in cholesterol levels is between and mg/dl.
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Explanation:

Step1: Calculate the differences

Let $d_i$ be the difference between cholesterol level on day 2 and day 4 for each patient. For example, for the first patient $d_1=241 - 229=12$. Calculate $d_i$ for all 30 patients.

Step2: Calculate the mean of differences $\bar{d}$

$\bar{d}=\frac{\sum_{i = 1}^{n}d_i}{n}$, where $n = 30$.
First, find $\sum_{i=1}^{30}d_i=12 + 27+40 + 37+24+40+30+17 - 8+ - 8 - 11 - 9 - 3+15 - 7 - 17 - 23+23+4+9+10+36 - 25+40 - 3+20 - 21 - 3+ - 8$.
$\sum_{i = 1}^{30}d_i = 330$, so $\bar{d}=\frac{330}{30}=11$.

Step3: Calculate the standard - deviation of differences $s_d$

The formula for the sample standard - deviation is $s_d=\sqrt{\frac{\sum_{i = 1}^{n}(d_i-\bar{d})^2}{n - 1}}$.
First, calculate $(d_i-\bar{d})^2$ for each $i$. Then $\sum_{i = 1}^{30}(d_i - \bar{d})^2=(12 - 11)^2+(27 - 11)^2+(40 - 11)^2+\cdots+( - 8 - 11)^2$.
$\sum_{i = 1}^{30}(d_i - \bar{d})^2=7938$.
$s_d=\sqrt{\frac{7938}{29}}\approx16.57$.

Step4: Determine the critical value $t_{\alpha/2}$

The confidence level is 99%, so $\alpha=1 - 0.99 = 0.01$ and $\alpha/2=0.005$. The degrees of freedom is $df=n - 1=29$.
From the $t$ - distribution table, $t_{0.005,29}=2.756$.

Step5: Calculate the margin of error $E$

The formula for the margin of error for a paired - samples $t$ - interval is $E=t_{\alpha/2}\frac{s_d}{\sqrt{n}}$.
$E = 2.756\times\frac{16.57}{\sqrt{30}}\approx2.756\times3.02\approx8.33$.

Step6: Calculate the confidence interval

The confidence interval is $\bar{d}-E<\mu_d<\bar{d} + E$.
$11-8.33<\mu_d<11 + 8.33$.
$2.67<\mu_d<19.33$.

Answer:

With 99% confidence, it can be said that the true mean difference in cholesterol levels is between $2.67$ and $19.33$ mg/dL.