Which of the following algebraic expressions can be used to express the statement "five less than y squared?" 5 − 2y 2y − 5 y2 −
1 answer:
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Based on the mean, standard deviation, and the graph, the indicated IQ score would be 115.
<h3>What is the indicated IQ score?</h3>First, find the z value:
P(Z<z) = 1 - tail value
= 1 - 0.1587
= 0.8413
According to the z-value, this gives a value of:
z = 1
The indicated IQ score is:
= mean + z value z standard deviation
= 100 + 1 x 15
= 115
Find out more on normal distributions at brainly.com/question/4079902
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Answer:
C) more than the absolute value of the critical value, we can conclude that the average monthly rate for an assisted-living facility is not equal to $3,300.
Step-by-step explanation:
Hello!
Interest hypothesizes is " the average monthly rate for one-bedroom assisted-living facility is equal to $3300" symbolically: μ = 3300
The study variable is:
X: Monthly rate for a one-bedroom assisted-living facility.
Since there is no information about the distribution of the variable, to be able to study the population mean, I'll assume that the variable has a normal distribution.
The hypothesis is:
H₀: μ = 3300
H₁: μ ≠ 3300
α: 0.05
The statistic to use, considering that there is no known population information and the sample size, is a Student t:
t= <u> X[bar] - μ </u> ~
S/√n
n= 12
X[bar]= $3690
S= $530
Using the sample data, calculate the statistic value:
t= <u> 3690 - 3300 </u> = 2.549
530/√12
The rejection region for this test is two-tailed, with critical values:
The decision rule is:
Reject the null hypothesis if t ≤ -2.201 or if t ≥ 2.201
Not reject the null hypothesis if -2.201 < t < 2.201
Since the calculated value (2.549) is greater than the right critical value (2.201) the decision is to reject the null hypothesis.
With a signification level of 5%, there is enough evidence to reject the null hypothesis. This means that the population mean of the monthly rate for a one-bedroom assisted living facility is different from $3300.
I hope it helps!