Question

Researchers fed mice a specific amount of Toxaphene, a poisonous pesticide, and studied their nervous systems to find out why Toxaphene causes tremors. The absolute refractory period, time required for nerves to recover after a stimulus, was measured and varies Normally. The measurements, in milliseconds, for four mice were 1.7, 1.8, 1.9, and 2.0.
Part A: Find the mean refractory period and the standard error of the mean.

Part B: Calculate a 90% confidence interval for the mean absolute refractory period for all mice when subjected to the same treatment.

Part C: Suppose the mean absolute refractory period for unpoisoned mice is known to be 1.3 milliseconds. Toxaphene poisoning should slow nerve recovery and therefore increase this period. Do the data give good evidence to support this theory? What can you conclude from a hypothesis test? Justify your response with statistical reasoning.

Answer 1

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.