Answer:
i think photomath will help you with those
Step-by-step explanation:
go to your appstore and download photomath
Answer:
![a)\ \ \bar x_m-\bar x_f=67.03\\\\b)\ \ E=15.7416\\\\c)\ \ CI=[51.2884, \ 82.7716]](https://tex.z-dn.net/?f=a%29%5C%20%5C%20%5Cbar%20x_m-%5Cbar%20x_f%3D67.03%5C%5C%5C%5Cb%29%5C%20%5C%20E%3D15.7416%5C%5C%5C%5Cc%29%5C%20%5C%20CI%3D%5B51.2884%2C%20%5C%2082.7716%5D)
Step-by-step explanation:
a. -Given that:

#The point estimator of the difference between the population mean expenditure for males and the population mean expenditure for females is calculated as:

Hence, the pointer is estimator 67.03
b. The standard error of the point estimator,
is calculated by the following following:

-And the margin of error, E at a 99% confidence can be calculated as:

Hence, the margin of error is 15.7416
c. The estimator confidence interval is calculated using the following formula:

#We substitute to solve for the confidence interval using the standard deviation and sample size values in a above:
![CI=\bar x_m-\bar x_f\ \pm z_{\alpha/2}\sqrt{\frac{\sigma_m^2}{n_m}+\frac{\sigma_f^2}{n_f}}\\\\=(135.67-68.64)\pm 15.7416\\\\=67.03\pm 15.7416\\\\=[51.2884, \ 82.7716]](https://tex.z-dn.net/?f=CI%3D%5Cbar%20x_m-%5Cbar%20x_f%5C%20%5Cpm%20z_%7B%5Calpha%2F2%7D%5Csqrt%7B%5Cfrac%7B%5Csigma_m%5E2%7D%7Bn_m%7D%2B%5Cfrac%7B%5Csigma_f%5E2%7D%7Bn_f%7D%7D%5C%5C%5C%5C%3D%28135.67-68.64%29%5Cpm%2015.7416%5C%5C%5C%5C%3D67.03%5Cpm%2015.7416%5C%5C%5C%5C%3D%5B51.2884%2C%20%5C%2082.7716%5D)
Hence, the 99% confidence interval is [51.2884,82.7716]
Answer:
P (T) = 1/4
P ( T | F ) = 1/2 = P(F)
The events are not independent.
Step-by-step explanation:
Let F the event of picking the white ball first
P (F)= 1/2 ( picking the white ball first)
Let T be the event of getting the white ball twice,
P (T) = P( getting white ball) * P( getting white ball)
=( 1/2)*(1/2)
= 1/4
Here P(T∩F) = P(T) because the probability of getting the white balls is the same as probability of getting the white ball first both the times.
P ( T | F ) = P (T∩F)/ P(F)
= (1/4)/ (1/2)
= (1/2)
= 1/2 = P(F)
For the events to be independent the conditional probability P ( T | F ) must be equal to P(T).
Hence the events are not independent.
Answer:
the answer is graph 1
Step-by-step explanation:
I think its $106.5 just add 17.75 6 times