Respuesta :
Answer:
We conclude that the brown trout's mean I.Q. is greater than four.
Step-by-step explanation:
We are given that the average brown trout's I.Q. is 4. Suppose you believe that the brown trout's mean I.Q. is greater than four.
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.
Let [tex]\mu[/tex] = the brown trout's mean I.Q.
So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] [tex]\leq[/tex] 4 {means that the brown trout's mean I.Q. is smaller than or equal to four}
Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] > 4 {means that the brown trout's mean I.Q. is greater than four}
The test statistics that would be used here One-sample t test statistics as we don't know about the population standard deviation;
T.S. = [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample mean I.Q. of brown tout = [tex]\frac{\sum X}{n}[/tex] = 4.92
s = sample standard deviation = [tex]\sqrt{\frac{\sum (X- \bar X)^{2} }{n-1} }[/tex] = 1.62
n = sample of brown trout = 12
So, the test statistics = [tex]\frac{4.92-4}{\frac{1.62}{\sqrt{12} } }[/tex] ~ [tex]t_1_1[/tex]
= 1.967
The value of t test statistics is 1.967.
Now, at 5% significance level the t table gives critical value of 1.796 at 11 degree of freedom for right-tailed test.
Since our test statistic is more than the critical value of t as 1.967 > 1.796, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.
Therefore, we conclude that the brown trout's mean I.Q. is greater than four.
