If ages are normally distributed then using a significance level of 0.05 , we can claim that the population mean is greater than 24.
Given sample mean of 25 years , sample size of 16 ,standard deviation of 2 years.
We have to find whether the population average is greater than 24.
We have to use t test because n is less than 30. It is right tailed.
We have to first form Hypothesis.
:μ>24,
:μ<24.
t=X-μ/S/
t critical at 5% significance level and degree of freedom=15 (16-1)=1.7531
t=24-25/2/
=-1/0.5
=-2
Because 1.7531 is greater than -2 so we will accept the null hypothesis.
Hence we can say that the average of population is greater than 24.
Learn more about t test at brainly.com/question/6589776
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Answer:
well if you subtract 29.5 minus 15.5 youn wil get 14
Step-by-step explanation:
Answer:
Quadrant 2
Step-by-step explanation:
Quadrant 1: Positive x and Positive y
Quadrant 2: Negative x and Positive y
Quadrant 3: Negative x and Negative y
Quadrant 4: Positive x and Negative y
For point (-4, 7), we have a Negative x and Positive y so it belongs in Quadrant 2.
I'd suggest you get rid of the fraction first. Mult. every term by 3. You will get
3y^2 + 10y + 3 = 0
From inspection, with coefficient a=3 and coeff. c = 3, the binomial factors could possibly begin with y or 3y: for example, y+1; also, the binom. factors may end in +1. Let's try the possible binomial factor 3y + 1.
Note that 10y separates into 9y+1y.
Then 3y^2 + 10y + 3 = 0 becomes
3y^2 + 9y + 1y + 3 = 0
Let's apply factoring by grouping:
3y^2 + 9y + 1y + 3 = 0
3y*(y + 3) + 1(y + 3) so y+3 is indeed a common factor.
Factoring y+3 out, we get (y+3)(3y + 1), which prove to be the correct set of factors. Multiply these together to ensure that the product is indeed y^2 + (10/3)y + 1.
#1:
Min 1
Max 64
Median 20.5
1st quartile 4
3rd quartile is 49
#2:
Min 2
Max 16
Median 9
1st quartile 4
3rd quartile 14
I know I did not find the mean for both of them. But I did everything else and I hope it helps you.
Hope this helps!
