Answer:
a. The mean refractory period= 1.85 and the standard error = 0.06455
b. 90% confidence interval for the mean absolute refractory period for all mice when subjected to the same treatment = 1.6981, 2.0019
c. Yes, the data give good evidence to support this theory
Step-by-step explanation:
a. The table below shows the calculations:
X (X-mean)^2
1.7 0.0225
1.8 0.0025
1.9 0.0025
2.0 0.0225
Total 7.4 0.05
Sample size: n=4
The mean is: \bar{x} = \frac{7.4}{4} = 1.85
The sample standard deviation, s = \sqrt{\frac{\sum \left ( x-\bar{x} \right )^{2}}{n-1}}=0.1291
The standard error, se= \frac{s}{\sqrt{n}}=\frac{0.1291}{2} = 0.06455
b. Degree of freedom: df = n-1 = 3
Critical value of t for 90% confidence interval is: 2.3534
The confidence interval is \bar{x}\pm t_{c}se = 1.85\pm 2.3534\cdot 0.06455=1.85\pm 0.1519 = (1.6981, 2.0019)
c. The Hypotheses are:
H_{0}:\mu=1.3,H_{1}:\mu>1.3
So the test statistics will be
t=\frac{\bar{x}-\mu}{s/\sqrt{n}}=8.52
The p-value is: 0.0017
We reject the null hypothesis because p-value is less than 0.05 . This indicates that the data gave good evidence to support this theory.
