Fthe coordinates of a and b are (-10, -3) and (7, 14), respectively, the x- intercept of is

Hello : let  A(-10,-3)    B(7,14)
the slope is :   (YB - YA)/(XB -XA)
(14+3)/17)  = 1
an equation is : y=ax+b     a is a slope

y = x +b

the line  F   through point  B(7,14) :  14 =7+b
b = 7
the equation is : y =X+7

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