Answer:
<em>H₀</em>: <em>μ</em>₁ = <em>μ</em>₂ vs, <em>Hₐ</em>: <em>μ</em>₁ > <em>μ</em>₂.
Step-by-step explanation:
A two-sample <em>z</em>-test can be performed to determine whether the claim made by the owner of pier 1 is correct or not.
It is provided that the weights of fish caught from pier 1 and pier 2 are normally distributed with equal population standard deviations.
The hypothesis to test whether the average weights of the fish in pier 1 is more than pier 2 is as follows:
<em>H₀</em>: The weights of fish in pier 1 is same as the weights of fish in pier 2, i.e. <em>μ</em>₁ = <em>μ</em>₂.
<em>Hₐ</em>: The weights of fish in pier 1 is greater than the weights of fish in pier 2, i.e. <em>μ</em>₁ > <em>μ</em>₂.
The significance level of the test is:
<em>α</em> = 0.05.
The test is defined as:

The decision rule for the test is:
If the <em>p</em>-value of the test is less than the significance level of 0.05 then the null hypothesis will be rejected and vice-versa.
Answer:
For 10 tosses we have that E(X)=10
Therefore E(i)= 1/10 +2/10 +3/10....10/10
This implies that 40/10=E(i)
Therefore E(10) =40/10
= 4.
Answer:
(5953.52,6046.49)
Step-by-step explanation:
We are given the following in the question:
Mean,
= 6,000 pounds
Sample size, n = 40
Alpha, α = 0.05
Standard deviation, σ = 150 pounds
95% Confidence interval:
Putting the values, we get,
are the limits within which 95% of the sample means occur.
The answer is that his average is 75.5 for the second nine weeks