Complete question :
A data set includes data from student evaluations of courses. The summary statistics are nequals92, x overbarequals4.09, sequals0.55. Use a 0.10 significance level to test the claim that the population of student course evaluations has a mean equal to 4.25. Assume that a simple random sample has been selected. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim.
Answer:
H0 : μ = 4.25
H1 : μ < 4.25
T = - 2.79
Pvalue =0.0026354
we conclude that there is enough evidence to conclude that population mean is different from 4.25 at 10%
Step-by-step explanation:
Given :
n = 92, xbar = 4.09, s = 0.55 ; μ = 4.25
H0 : μ = 4.25
H1 : μ < 4.25
The test statistic :
T = (xbar - μ) ÷ s / √n
T = (4.09 - 4.25) ÷ 0.55/√92
T = - 0.16 / 0.0573414
T = - 2.79
The Pvalue can be obtained from the test statistic, using the Pvalue calculator
Pvalue : (Z < - 2.79) = 0.0026354
Pvalue < α ; Hence, we reject the Null
Thus, we conclude that there is enough evidence to conclude that population mean is different from 4.25 at 10%
<h3>
Answer: Choice C</h3>
Work Shown:
In the third step, I used the rule that 
. You subtract the exponents when dividing exponents of the same base.
After that, I used the rule 
which is the same as saying 
to end up with choice C as the answer.
Answer:
N = 4
Step-by-step explanation:
In the equation 2n + 7 = 15, we have to isolate the variable. But, before that, we have to get rid of anything else. Like the +7 in the equation. So we can subtract from each side.
2n + 7 = 15
Take out the +7 by subtracting 7 on both sides.
2n + 7 - 7 = 15 - 7
2n = 8
Now, to get rid of the 2 next to the n, we have to divide both sides by 2
2n/2 = 8/2
n=4
102x3=306
just multiply it
