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%
2.12=2.120 that is the answer
The inequalities and the points are
- y + 2 ≤ 5x: (5, -2) and (10, 1)
- y + x > -4: (5, -2) and (10, 1)
- y ≤ 1/3x: (5, -2) and (10, 1))
<h3>How to graph the
system of inequalities?</h3>
The system of inequalities is given as:
y + 2 ≤ 5x
y + x > -4
y ≤ 1/3x
Next, we plot each inequality in the system of inequalities using a graphing calculator
See attachment for the graph of the system of inequalities y + 2 ≤ 5x, y + x > -4 and y ≤ 1/3x
From the attached graph, we have the following points:
y + 2 ≤ 5x: (5, -2) and (10, 1)
y + x > -4: (5, -2) and (10, 1)
y ≤ 1/3x: (5, -2) and (10, 1)
Read more about inequalities at:
brainly.com/question/24372553
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