In one day, 125,683 people fly in and out of an airport in texas. how many thousands of people fly in and out that day?

125 thousands of people fly in and out that day.    

Step-by-step explanation:

Given : In one day, 125,683 people fly in and out of an airport in Texas.

To find : How many thousands of people fly in and out that day?

Solution :

Using the place value system,

Hundred Th.   Ten Th.   Thousand   Hundred    Tens    Ones

       1                    2                5                6              8           3

There are 125 thousands 6 hundred and 83.

As in one day, 125,683 people fly in and out of an airport in Texas.

Number of thousands of people fly in and out that day is 125.

Therefore, 125 thousands of people fly in and out that day.

We're told that "the claim that the population of student course evaluations has a mean equal to 3.50". So this means μ=3.50 makes up the null H0

The alternative would be H1: μ ≠ 3.50 since it's the opposite of the claim made in the null.

We go with answer choice D to form the null and alternative hypotheses.

The sign ≠ in the alternative hypothesis tell us that we have a two tail test.

Let's compute the test statistic

z = (xbar - mu)/(s/sqrt(n))

z = (3.44 - 3.50)/(0.67/sqrt(89))

z = -0.84483413122896

z = -0.84

The test statistic is roughly -0.84

Despite not knowing what sigma is (aka the population standard deviation), we can see that n > 30 is the case. So we can use the Z distribution. This is the standard normal distribution. When n > 30, the T distribution is fairly approximately the same as the Z distribution.

Use a calculator or a Z table to determine that

P(Z < -0.84) = 0.2005

which is approximate

Because we're doing a two-tail test, this means we double that result to get 2*0.2005 = 0.401

The p-value is roughly 0.401

Since the p-value is larger than alpha = 0.05, we don't have enough evidence to reject the null. So you can say that we fail to reject the null, or we accept the null.

The conclusion based on that means that μ=3.50  must be true (unless other evidence comes along to disprove this). In other words, the mean evaluation score from students appears to be 3.50