Bar graphs sorry if I’m not right
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
He worked 5 hours in total: 35$ for the first hour, + 23 + 23 + 23 + 23 = 127$ for five hours of work.
(2,5)(-1,8)
slope = (8 - 5) / (-1 - 2) = 3/-3 = -1
y - y1 = m(x - x1)
slope(m) = -1
(2,5)...x1 = 2 and y1 = 5
now we sub
y - 5 = -1(x - 2) <==
Answer:
the answer is B
Step-by-step explanation:
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