Answer:
Explanation:
Considering the brown tout,
The mean of the set of data given is
Mean = (5 + 4 + 7 + 3 + 6 + 4 + 5 + 3 + 6 + 3 + 8 + 5)/12 = 4.92
Standard deviation = √(summation(x - mean)²/n
n = 12
Summation(x - mean)² = (5 - 4.92)^2 + (4 - 4.92)^2 + (7 - 4.92)^2 + (3 - 4.92)^2 + (6 - 4.92)^2 + (4 - 4.92)^2 + (5 - 4.92)^2 + (3 - 4.92)^2 + (6 - 4.92)^2 + (3 - 4.92)^2 + (8 - 4.92)^2 + (5 - 4.92)^2 = 28.9168
Standard deviation = √(28.9168/12) = 1.55
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,
µ ≥ 4
For the alternative hypothesis,
µ < 4
This is a left 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 = 12,
Degrees of freedom, df = n - 1 = 12 - 1 = 11
t = (x - µ)/(s/√n)
Where
x = sample mean = 4.92
µ = population mean = 4
s = samples standard deviation = 1.55
t = (4.92 - 4)/(1.55/√12) = 2.06
We would determine the p value using the t test calculator. It becomes
p = 0.98
Assuming a significance level of 5%
Since alpha, 0.05 < than the p value, 0.98, then we would fail to reject the null hypothesis. Therefore, At a 5% level of significance, the sample data showed significant evidence that the brown trout’s mean I.Q. is greater than 4