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%
X ・y^5
x times y to the fifth
C both median and mode are 7
Answer:
Step-by-step explanation:
Given
- A is between C and T
- CA =4x + 6
- AT =7x-6
- CT =132
As per segment addition postulate:
- CT = CA + AT
- 4x + 6 + 7x - 6 = 132
- 11x = 132
- x = 132/11
- x = 12
