This question is incomplete, the complete question is;
Assume that two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Which distribution is used to test the claim that women have a higher mean resting heart rate than men?
A.t
B.F
C.Normal
D. Chi-square
Answer:
A) t test
Step-by-step explanation:
A t-test uses sample information to assess how plausible it is for the population means μ1 and μ2 to be equal.
The formula for a t-statistic for two population means (with two independent samples), with unknown population variances shows us how to calculate t-test with mean and standard deviation and it depends on the assumption of having an equal variance or not.
If the variances are assumed to be not equal,
he formula is:
t = (bar X₁ - bar X₂) / √( s₁²/n₁ + s₂²/n₂ )
If the variances are assumed to be equal, the formula is:
t = (bar X₁ - bar X₂) / √ (((n₁ - 1)s₁² + (n₂ - 1)s₂²) / (n₁ + n₂ - 2)) ( 1/n₁ + 1/n₂)
it called t-test for independent samples because the samples are not related to each other, in a way that the outcomes from one sample are unrelated from the other sample.
hence option A is correct. t test
0.004 kilometers... so was there a typo?
S^2 = one face of a cube
S^2 × 6 = entire Surface Area (SA) of cube
One face = 9^2 = 81 cm^2
SA = 81 × 6 = 486 cm^2
The answer is D.
|-4| < 4
4 < 4
|-1| < 0
1 < 0
|-6| = -6
6 = -6
|-14| > 10
14 > 10
What you want is P(6∩1) or P(1∩6) or P(2∩5) or P(5∩2) or P(3∩4) or P(4∩3).
The events of rolling the dice are independent (i.e. they don't affect one another) so:
E.g.
P(6∩1) = P(6) * P(1)
P(2∩5) = P(2) * P(5)
The probability of getting a given number on a roll is 1/6 for both dice.
So:
P(6∩1) = 1/6 * 1/6 = 1/36
This is the same for any arrangement of numbers you could get from rolling two dice.
So, we can see that there are 6 arrangements of numbers that will give a sum of 7 and so that is 6 * 1/36 = 6/36 = 1/6
