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
Answer:
(x-7) (x^2-5)
Step-by-step explanation:
x^3 -7x^2 -5x+35
Make 2 groups
x^3 -7x^2 -5x+35
Factor x^2 from the first group and -5 from the second group
x^2 (x-7) -5(x-7)
Now factor (x-7) out
(x-7) (x^2-5)
You have a 1/2 chance of spinning an even number on the spinner, and a 1/2 chance of rolling an odd number on the die, 1/2 times 1/2 give you a 1/4 chance of doing both of those tasks.
Answer:
Henry will get $15.77 back.
Step-by-step explanation:
20 - 4.23 = 15.77
