Suppose the weights of fish caught from pier 1 and pier 2 are normally distributed with equal population standard deviations. The natural assumption is that the mean weights of fish caught at the two piers are equal. The owner of pier 1 thinks that the average at his pier is greater. To test at the 5% level of significance that the average weights of the fish in pier 1 is more than pier 2 the null and the alternative hypotheses are

H₀: μ₁ = μ₂ vs, Hₐ: μ₁ > μ₂.

Step-by-step explanation:

A two-sample z-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:

H₀: The weights of fish in pier 1 is same as the weights of fish in pier 2, i.e. μ₁ = μ₂.

Hₐ: The weights of fish in pier 1 is greater than the weights of fish in pier 2, i.e. μ₁ > μ₂.

The significance level of the test is:

α = 0.05.

The test is defined as:

The decision rule for the test is:

If the p-value of the test is less than the significance level of 0.05 then the null hypothesis will be rejected and vice-versa.

7 weeks

step-by-step explanation:

convert each fraction to the least common multiple of 20, so 3/4 = 15/20 & 1/10 = 2/20, then divide 15 by 2 and you get a little over 7, so the food will last 7 weeks.

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