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.
First you plug in the information you already know. 163.5= 
Now you have to solve for the mass so you multiply 0.492 by 163.5 
Lastly, you simplify and you will get 80.44. Hope this helped :)