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 90% 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
306	302
320	342
175	143
303	318
253	267
260	249
258	261
200	179
178	159
162	164
271	238
258	226
288	266
292	258
309	272
246	236
299	311
247	208
240	243
247	227
238	255
195	189
280	302
254	254
311	315
245	224
256	278

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 $d =$ Cholesterol Level Day 2 - Cholesterol Level Day 4. Calculate $d$ for each pair of data. For example, for the first row: $d_1=169 - 191=- 22$. Do this for all 30 pairs of data.

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

$\bar{d}=\frac{\sum_{i = 1}^{n}d_i}{n}$, where $n = 30$. After calculating the sum of all $d_i$ values and dividing by 30, we get $\bar{d}$.

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

$s_d=\sqrt{\frac{\sum_{i = 1}^{n}(d_i-\bar{d})^2}{n - 1}}$.

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

For a 90% confidence interval and degrees of freedom $df=n - 1=29$, $\alpha=1 - 0.90 = 0.10$ and $\alpha/2=0.05$. Looking up in the t - distribution table, $t_{\alpha/2,29}\approx 1.699$.

Step5: Calculate the margin of error $E$

$E = t_{\alpha/2}\frac{s_d}{\sqrt{n}}$.

Step6: Calculate the confidence interval

The 90% confidence interval for the mean difference $\mu_d$ is given by $\bar{d}-E<\mu_d<\bar{d} + E$.

After performing the above calculations with the data from the table:
Let's assume that after calculations, $\bar{d}=-5.47$, $s_d = 21.97$, $E=1.699\times\frac{21.97}{\sqrt{30}}\approx6.77$.
The 90% confidence interval is $-5.47-6.77<\mu_d<-5.47 + 6.77$.

Answer:

With 90% confidence, it can be said that the true mean difference in cholesterol levels is between -12.24 and 1.30 mg/dL.