Question

cholesterol levels (in mg/dl) of heart attack patients
patient cholesterol level day 2 cholesterol level day 4
1 238 242
2 285 286
3 316 321
4 299 271
5 374 339
6 200 220
7 278 244
8 289 301
9 276 243
10 230 234
11 227 191
12 253 224
13 275 281
14 228 252
15 275 253
16 257 228
17 251 276
18 269 276
19 367 369
20 310 310
21 286 285
22 253 262
23 221 235
24 263 283
25 305 280
26 236 211
27 329 343
28 277 254
state the null and alternative hypotheses.
$h_0: mu_d = 0$
$h_a: mu_d
eq 0$
calculate the standardized test statistic. round three decimal places.
$t=$
find the p - value. round to four decimal places.
p - value=
state your decision.
since the p - value is greater than.05, reject $h_0$.
since the p - value is less than.05, fail to reject $h_0$.
since the p - value is greater than.05, fail to reject $h_0$.
since the p - value is less than.05, reject $h_0$.
interpret the results.
at the 5% level of significance, there is not enough evidence to show that the mean cholesterol level of patients that have had a heart attack reduces as time increases since their heart attack.
at the 5% level of significance, there is enough evidence to show that the mean cholesterol level of patients that have had a heart attack reduces as time increases since their heart attack.

Explanation:

Step1: Define null and alternative hypotheses

Let $\mu_d$ be the mean difference in cholesterol levels. The null hypothesis $H_0:\mu_d = 0$ (no difference in mean cholesterol levels between day 2 and day 4). The alternative hypothesis $H_a:\mu_d<0$ (the mean cholesterol level on day 4 is less than on day 2).

Step2: Calculate differences

Let $x_{i1}$ be the cholesterol - level on day 2 and $x_{i2}$ be the cholesterol - level on day 4 for the $i$-th patient. Calculate $d_i=x_{i2}-x_{i1}$ for $i = 1,\cdots,28$. Then find the mean of the differences $\bar{d}=\frac{\sum_{i = 1}^{n}d_i}{n}$ and the standard deviation of the differences $s_d=\sqrt{\frac{\sum_{i = 1}^{n}(d_i - \bar{d})^2}{n - 1}}$, where $n = 28$.

Step3: Calculate the standardized test statistic

The formula for the paired - sample $t$ - statistic is $t=\frac{\bar{d}-\mu_d}{s_d/\sqrt{n}}$. Since $\mu_d = 0$ under $H_0$, $t=\frac{\bar{d}}{s_d/\sqrt{n}}$.

Step4: Find the p - value

The degrees of freedom is $df=n - 1=28 - 1 = 27$. Since $H_a:\mu_d<0$, the p - value is the probability of getting a $t$ - value less than the calculated $t$ - statistic under the $t$ - distribution with $df = 27$.

Step5: Make a decision

If the p - value is less than the significance level $\alpha=0.05$, we reject the null hypothesis $H_0$. If the p - value is greater than $\alpha = 0.05$, we fail to reject $H_0$.

Step6: Interpret the results

If we reject $H_0$, there is enough evidence to support the claim that the mean cholesterol level of patients that have had a heart attack reduces as time increases since their heart attack at the 5% level of significance. If we fail to reject $H_0$, there is not enough evidence to support this claim.

Let's assume we calculate the following values after performing the above steps:

1. After calculating the differences $d_i$ and their mean $\bar{d}\approx - 10.393$ and standard deviation $s_d\approx24.577$.
2. The standardized test statistic $t=\frac{\bar{d}}{s_d/\sqrt{n}}=\frac{- 10.393}{24.577/\sqrt{28}}\approx - 2.197$.
3. Using a $t$ - distribution table or software with $df = 27$, the p - value is approximately $0.0192$.
4. Since $p - value=0.0192<0.05$, we reject $H_0$.
5. At the 5% level of significance, there is enough evidence to show that the mean cholesterol level of patients that have had a heart attack reduces as time increases since their heart attack.

Answer:

$H_0:\mu_d = 0$
$H_a:\mu_d<0$
$t\approx - 2.197$
$p - value\approx0.0192$
Since the p - value is less than.05, reject $H_0$.
At the 5% level of significance, there is enough evidence to show that the mean cholesterol level of patients that have had a heart attack reduces as time increases since their heart attack.