Question

suppose you are a researcher in a hospital. you are experimenting with a new sedative. you collect data from a random sample of 9 patients. the period of effectiveness of the sedative for each patient (in hours) is as follows. click this link download csv to download the data. a) what is a point estimate for the population mean length of time? (round answer to 4 decimal places) b) what must be true in order to construct a confidence interval in this situation? c) construct a 90% confidence interval for the population mean length of time. enter your answer as an open - interval (i.e., parentheses example (5.2314,8.1245)) round upper and lower bounds to 4 decimal places.

Explanation:

Step1: Calculate the sample mean (point - estimate)

The formula for the sample mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\).
Here, \(n = 9\) and \(x=\{2.6,2.6,3,2.1,2.6,2.8,2.1,2.7,3\}\)
\(\sum_{i=1}^{9}x_{i}=2.6 + 2.6+3 + 2.1+2.6+2.8+2.1+2.7+3=23.5\)
\(\bar{x}=\frac{23.5}{9}\approx2.6111\)

Step2: Calculate the sample standard deviation \(s\)

The formula for the sample standard deviation \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}\)
\((x_{1}-\bar{x})=(2.6 - 2.6111)=- 0.0111\), \((x_{2}-\bar{x})=(2.6 - 2.6111)=-0.0111\), \((x_{3}-\bar{x})=(3 - 2.6111)=0.3889\), \((x_{4}-\bar{x})=(2.1 - 2.6111)=-0.5111\), \((x_{5}-\bar{x})=(2.6 - 2.6111)=-0.0111\), \((x_{6}-\bar{x})=(2.8 - 2.6111)=0.1889\), \((x_{7}-\bar{x})=(2.1 - 2.6111)=-0.5111\), \((x_{8}-\bar{x})=(2.7 - 2.6111)=0.0889\), \((x_{9}-\bar{x})=(3 - 2.6111)=0.3889\)
\(\sum_{i = 1}^{9}(x_{i}-\bar{x})^{2}=(-0.0111)^{2}+(-0.0111)^{2}+(0.3889)^{2}+(-0.5111)^{2}+(-0.0111)^{2}+(0.1889)^{2}+(-0.5111)^{2}+(0.0889)^{2}+(0.3889)^{2}\)
\(=0.00012321+0.00012321 + 0.15124321+0.26122321+0.00012321+0.03568321+0.26122321+0.00790321+0.15124321\)
\(=0.868895\)
\(s=\sqrt{\frac{0.868895}{9 - 1}}=\sqrt{\frac{0.868895}{8}}\approx0.3289\)

Step3: Find the \(t\) - value

For a 90% confidence interval and \(n-1=9 - 1 = 8\) degrees of freedom, from the \(t\) - distribution table, \(t_{\alpha/2}=1.8595\)

Step4: Calculate the margin of error \(E\)

The formula for the margin of error \(E=t_{\alpha/2}\frac{s}{\sqrt{n}}\)
\(E = 1.8595\times\frac{0.3289}{\sqrt{9}}=1.8595\times\frac{0.3289}{3}\approx0.2057\)

Step5: Calculate the confidence interval

The confidence interval for the population mean \(\mu\) is \(\bar{x}-E<\mu<\bar{x}+E\)
\(2.6111-0.2057 <\mu<2.6111 + 0.2057\)
\(2.4054<\mu<2.8168\)

Answer:

a) \(2.6111\)
b) The population from which the sample is drawn must be approximately normally distributed.
c) \((2.4054,2.8168)\)