The following are the systolic blood pressures (mm Hg) of 12 patients undergoing drug therapy for hypertension: 183, 152, 178, 157, 194, 163, 144, 114, 178, 152, 118, 158. Can we conclude, based on these data, that the population mean is less than 165? Let a = 0.05. What assumptions are necessary?
The Correct Answer and Explanation is:
We are tasked with testing whether the population mean systolic blood pressure of the patients undergoing drug therapy for hypertension is less than 165, using a significance level of α = 0.05.
To test this, we will conduct a one-sample t-test. The null hypothesis (H₀) and the alternative hypothesis (H₁) are as follows:
- H₀ (null hypothesis): The population mean systolic blood pressure is equal to 165 (μ = 165).
- H₁ (alternative hypothesis): The population mean systolic blood pressure is less than 165 (μ < 165).
Assumptions:
- Independence: The observations (systolic blood pressures) are independent of one another, meaning that each patient’s blood pressure is not influenced by others.
- Normality: The data should be approximately normally distributed. Since the sample size is small (n = 12), we cannot rely on the Central Limit Theorem, so we must assume the data is roughly normal or check for any significant outliers.
Test Statistic:
We use the formula for the t-statistic: t=xˉ−μ0snt = \frac{\bar{x} – \mu_0}{\frac{s}{\sqrt{n}}}t=nsxˉ−μ0
Where:
- xˉ\bar{x}xˉ is the sample mean,
- μ0\mu_0μ0 is the hypothesized population mean (165),
- sss is the sample standard deviation, and
- nnn is the sample size.
Calculation:
- Sample Mean (xˉ\bar{x}xˉ):
xˉ=183+152+178+157+194+163+144+114+178+152+118+15812=183912=153.25\bar{x} = \frac{183 + 152 + 178 + 157 + 194 + 163 + 144 + 114 + 178 + 152 + 118 + 158}{12} = \frac{1839}{12} = 153.25xˉ=12183+152+178+157+194+163+144+114+178+152+118+158=121839=153.25
- Sample Standard Deviation (s):
First, calculate the squared differences from the mean and then find the square root of the average:
s=∑(xi−xˉ)2n−1≈28.33s = \sqrt{\frac{\sum (x_i – \bar{x})^2}{n – 1}} \approx 28.33s=n−1∑(xi−xˉ)2≈28.33
- t-Statistic:
t=153.25−16528.3312≈−11.758.18≈−1.44t = \frac{153.25 – 165}{\frac{28.33}{\sqrt{12}}} \approx \frac{-11.75}{8.18} \approx -1.44t=1228.33153.25−165≈8.18−11.75≈−1.44
Degrees of Freedom:
The degrees of freedom for this test is n−1=12−1=11n – 1 = 12 – 1 = 11n−1=12−1=11.
Critical Value:
For a one-tailed test with α = 0.05 and 11 degrees of freedom, the critical t-value can be found using a t-distribution table. The critical value is approximately -1.796 (since it’s a one-tailed test, we look at the lower tail).
Conclusion:
Since the calculated t-statistic (-1.44) is greater than the critical value (-1.796), we fail to reject the null hypothesis. There is insufficient evidence at the 0.05 significance level to conclude that the population mean systolic blood pressure is less than 165.
Final Answer:
We cannot conclude, based on these data, that the population mean systolic blood pressure is less than 165 at the 0.05 significance level. The assumptions necessary for this test are the independence of observations and the normality of the data.
