Randy filled out a deposit slip today. He deposited two checks for $123.17 and $34.68 and took out $50 in cash.
2 answers:
You might be interested in
Answer:
Therefore the 95% confidence interval is (25,707.480 < E < 26,744.920)
Step-by-step explanation:
n = 77
mean u = 26,226.2 bushels per acre
standard deviation s = 2,322.32
let E = true mean
let A = test statistic
Find 95% Confidence Interval
so
let A = (u - E) * ( / s) be the test statistic
we want P( average_l < A < average_u ) = 95%
look for lower 2.5% and the upper 97.5% Because I think this is a 2-tail test
average_l = -1.96 which corresponds to the 2.5%
average_u = 1.96
P( -1.96 < A < 1.96) = 95%
P( -1.96 < (u - E) * ( / s) < 1.96) = 95%
Solve for the true mean E ok
E < u + 1.96* (s / )
from -1.96 < (u - E) * ( / s)
E < 26,226.2 + 1.96*( 2,322.32 / )
E < 26,226.2 + 1.96*( 2,322.32 / )
E < 26,226.2 + 518.7197348105429466
upper bound is 26,744.9197
or
u - 1.96* (s / ) < E
26,226.2 - 518.7197348105429466 < E
25,707.48026519 < E
lower bound is 25,707.48026519
Therefore the 95% confidence interval is (25,707.480 < E < 26,744.920)
Answer:
The reason why standard deviation of the entire class is greater than standard deviation of males and females considered separately, is that mean values for males and females are different from each other.
Step-by-step explanation:
The concept of mean is well represented by the following formula
mean = , where x1, x2, xn are the observations and N is the number of observations (population).
Standard deviation represents the distance between each observation and the mean of the population (all observations). The formula for this parameter is:
Standard deviation =√[((x1 - x)² + (x2-x)² + ....+ (xn-x)²)/N-1], where x1, x2,..., xn are the observations and x is the mean value.
In this case you have that each height registered is an observation and the number of observations represents the N value. As you can see if the mean for males is different from that of females their standard deviation will be different too. Usually males have heigths greater than that of females (1.77 vs 1.64, in USA for example), and heights inside each group will be more similar than between groups. Then, when you mix all observation there will be an increase in standard deviation, because you are mixing very different heigths