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An employee at an aquarium monitors how much their sea otters eat. The amount of food a particular otter eats
daily is approximately normally distributed with a mean of 17 pounds and a standard deviation of 1 pound. They
suspected this otter was not eating enough, so they took a random sample n = 10 days and observed a sample
mean of T = 16.5 pounds of food per day.
To see how likely a sample like this was to occur by random chance alone, the employee performed a simulation.
They simulated 40 samples of n = 10 values from a normal population with a mean of 17 pounds and a
standard deviation of 1 pound. They recorded the mean of the values in each sample. Here are the sample
means from their 40 samples:


Sagot :

Answer:

0.125

Step-by-step explanation:

5/40 simulated results were at or below mean 16.5

The approximate p-value of the test that's given in this scenario will be p-value = 0.125.

What is p-value?

P-value simply means the probability that a test statistic can be seen to be extreme or more.

From the complete question, the approximate p-value of the test is 0.125. The p-value measures the probability that a difference could have taken place by random chance.

In conclusion, the observed difference in this case can be attributed to chance by 12.5%.

Learn more p-value on:

https://brainly.com/question/13786078

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