Answer:
Option 3. The researchers can conclude that the fish are smaller than what is normal because the difference between 12.3 inches and 11.2 inches is much larger than the expected sampling error.
Explanation:
We are given the following in the question:
Population mean, μ = 12.3 inches
Sample mean,
= 11.2 inches
Sample size, n = 100
Alpha, α = 0.05
Population standard deviation, σ = 3 inches
First, we design the null and the alternate hypothesis
![H_(0): \mu = 12.3\text{ inches}\\H_A: \mu < 12.3\text{ inches}](https://img.qammunity.org/2021/formulas/mathematics/college/xov0gq8wsalqgugglgp0u5oi47acc677w8.png)
We use one-tailed z test to perform this hypothesis.
Formula:
![z_(stat) = \displaystyle\frac{\bar{x} - \mu}{(\sigma)/(√(n)) }](https://img.qammunity.org/2021/formulas/mathematics/college/b58owkon7rcyqrb9ldz6ydgwjck1hf8bev.png)
Putting all the values, we have
![z_(stat) = \displaystyle(11.2 -12.3)/((3)/(√(100)) ) = -3.67](https://img.qammunity.org/2021/formulas/mathematics/college/iildn6z2xj35xubyphonts0plmg4a0nswv.png)
Now,
![z_(critical) \text{ at 0.05 level of significance } = -1.64](https://img.qammunity.org/2021/formulas/mathematics/college/n8u0wqkxb2ucncs6fztzhw40j724m64ra4.png)
Since,
![z_(stat) < z_(critical)](https://img.qammunity.org/2021/formulas/mathematics/college/r5o6oexg3zrsx7zj5x5cgc2v73cma5hw4t.png)
We reject the null hypothesis and accept the alternate hypothesis.
Thus, we conclude that this year they are catching smaller than usual largemouth bass.
Option 3. The researchers can conclude that the fish are smaller than what is normal because the difference between 12.3 inches and 11.2 inches is much larger than the expected sampling error.