Aresearcher wants to see if there is a difference in the number of emplyees at a company scheduled to work on a specific bridge. at alpha = .05, is there a difference in the average number of employees working at each bridge?

STEP BY STEP explanation: The hypotheses to test here would be:

Null hypothesis: there is no difference in the average number of employees working at each bridge

Alternative hypothesis: there is a difference in the average number of employees working at each bridge

A significance level or alpha of 0.05 indicates a 5% risk of concluding that a difference exists when there is no actual difference. That is, 0.05 is the probability of rejecting the null hypothesis when it is true.

Fix in the necessary values in the question such as the number of observations, the mean, the standard error of the mean, the standard deviant of the mean.

In a probability distribution plot, map out the two critical regions in order to get your answer. There are no values here in your question.

3030 ur welcome

step-by-step explanation:

1/3

step-by-step explanation:

to find this, you can divide 12 by 4 which is 3 or you can multiply 4 until you reach 12 (4x3=12). both should give you 3, meaning theres three times less bananas than there are oranges.