Answer:
Step-by-step explanation:
Part A
Mean = (2.2 + 2.4 + 2.5 + 2.5 + 2.6 + 2.7)/6 = 2.48
Standard deviation = √(summation(x - mean)²/n
n = 6
Summation(x - mean)² = (2.2 - 2.48)^2 + (2.4 - 2.48)^2 + (2.5 - 2.48)^2 + (2.5 - 2.48)^2 + (2.6 - 2.48)^2 + (2.7 - 2.48)^2 = 0.1484
Standard deviation = √(0.1484/6
s = 0.16
Standard error = s/√n = 0.16/√6 = 0.065
Part B
Confidence interval is written as sample mean ± margin of error
Margin of error = z × s/√n
Since sample size is small and population standard deviation is unknown, z for 98% confidence level would be the t score from the student t distribution table. Degree of freedom = n - 1 = 6 - 1 = 5
Therefore, z = 3.365
Margin of error = 3.365 × 0.16/√6 = 0.22
Confidence interval is 2.48 ± 0.22
Part C
We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean
For the null hypothesis,
H0: µ = 2.3
For the alternative hypothesis,
H1: µ > 2.3
This is a right tailed test
Since the number of samples is small and no population standard deviation is given, the distribution is a student's t.
Since n = 6
Degrees of freedom, df = n - 1 = 6 - 1 = 5
t = (x - µ)/(s/√n)
Where
x = sample mean = 2.48
µ = population mean = 2.3
s = samples standard deviation = 0.16
t = (2.48 - 2.3)/(0.16/√6) = 2.76
We would determine the p value using the t test calculator. It becomes
p = 0.02
Assuming significance level, alpha = 0.05.
Since alpha, 0.05 > than the p value, 0.02, then we would reject the null hypothesis. Therefore, At a 5% level of significance, the sample data showed significant evidence that the mean absolute refractory period for all mice when subjected to the same treatment increased.
