A Nissan Motor Corporation advertisement read, "The average man’s I. Q. is 107. The average brown trout’s I. Q. is 4. So why can’t man catch brown trout?" Suppose you believe that the brown trout’s mean I. Q. is greater than 4. You catch 12 brown trout. A fish psychologist determines the I. Q.s as follows: 5; 4; 7; 3; 6; 4; 5; 3; 6; 3; 8; 5. Conduct a hypothesis test of your belief.

Step-by-step 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

the rate of change is 25