It is the last numbers I'm positive
Answer:
Using a formula, the standard error is: 0.052
Using bootstrap, the standard error is: 0.050
Comparison:
The calculated standard error using the formula is greater than the standard error using bootstrap
Step-by-step explanation:
Given
Sample A Sample B
![x_B = 50](https://tex.z-dn.net/?f=x_B%20%3D%2050)
![n_B =250](https://tex.z-dn.net/?f=n_B%20%3D250)
Solving (a): Standard error using formula
First, calculate the proportion of A
![p_A = \frac{x_A}{n_A}](https://tex.z-dn.net/?f=p_A%20%3D%20%5Cfrac%7Bx_A%7D%7Bn_A%7D)
![p_A = \frac{30}{100}](https://tex.z-dn.net/?f=p_A%20%3D%20%5Cfrac%7B30%7D%7B100%7D)
![p_A = 0.30](https://tex.z-dn.net/?f=p_A%20%3D%200.30)
The proportion of B
![p_B = \frac{x_B}{n_B}](https://tex.z-dn.net/?f=p_B%20%3D%20%5Cfrac%7Bx_B%7D%7Bn_B%7D)
![p_B = \frac{50}{250}](https://tex.z-dn.net/?f=p_B%20%3D%20%5Cfrac%7B50%7D%7B250%7D)
![p_B = 0.20](https://tex.z-dn.net/?f=p_B%20%3D%200.20)
The standard error is:
![SE_{p_A-p_B} = \sqrt{\frac{p_A * (1 - p_A)}{n_A} + \frac{p_A * (1 - p_B)}{n_B}}](https://tex.z-dn.net/?f=SE_%7Bp_A-p_B%7D%20%3D%20%5Csqrt%7B%5Cfrac%7Bp_A%20%2A%20%281%20-%20p_A%29%7D%7Bn_A%7D%20%2B%20%5Cfrac%7Bp_A%20%2A%20%281%20-%20p_B%29%7D%7Bn_B%7D%7D)
![SE_{p_A-p_B} = \sqrt{\frac{0.30 * (1 - 0.30)}{100} + \frac{0.20* (1 - 0.20)}{250}}](https://tex.z-dn.net/?f=SE_%7Bp_A-p_B%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B0.30%20%2A%20%281%20-%200.30%29%7D%7B100%7D%20%2B%20%5Cfrac%7B0.20%2A%20%281%20-%200.20%29%7D%7B250%7D%7D)
![SE_{p_A-p_B} = \sqrt{\frac{0.30 * 0.70}{100} + \frac{0.20* 0.80}{250}}](https://tex.z-dn.net/?f=SE_%7Bp_A-p_B%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B0.30%20%2A%200.70%7D%7B100%7D%20%2B%20%5Cfrac%7B0.20%2A%200.80%7D%7B250%7D%7D)
![SE_{p_A-p_B} = \sqrt{\frac{0.21}{100} + \frac{0.16}{250}}](https://tex.z-dn.net/?f=SE_%7Bp_A-p_B%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B0.21%7D%7B100%7D%20%2B%20%5Cfrac%7B0.16%7D%7B250%7D%7D)
![SE_{p_A-p_B} = \sqrt{0.0021+ 0.00064}](https://tex.z-dn.net/?f=SE_%7Bp_A-p_B%7D%20%3D%20%5Csqrt%7B0.0021%2B%200.00064%7D)
![SE_{p_A-p_B} = \sqrt{0.00274}](https://tex.z-dn.net/?f=SE_%7Bp_A-p_B%7D%20%3D%20%5Csqrt%7B0.00274%7D)
![SE_{p_A-p_B} = 0.052](https://tex.z-dn.net/?f=SE_%7Bp_A-p_B%7D%20%3D%200.052)
Solving (a): Standard error using bootstrapping.
Following the below steps.
- Open Statkey
- Under Randomization Hypothesis Tests, select Test for Difference in Proportions
- Click on Edit data, enter the appropriate data
- Click on ok to generate samples
- Click on Generate 1000 samples ---- <em>see attachment for the generated data</em>
From the randomization sample, we have:
Sample A Sample B
![x_B = 57](https://tex.z-dn.net/?f=x_B%20%3D%2057)
![n_B =250](https://tex.z-dn.net/?f=n_B%20%3D250)
![p_A = 0.228](https://tex.z-dn.net/?f=p_A%20%3D%200.228)
So, we have:
![SE_{p_A-p_B} = \sqrt{\frac{p_A * (1 - p_A)}{n_A} + \frac{p_A * (1 - p_B)}{n_B}}](https://tex.z-dn.net/?f=SE_%7Bp_A-p_B%7D%20%3D%20%5Csqrt%7B%5Cfrac%7Bp_A%20%2A%20%281%20-%20p_A%29%7D%7Bn_A%7D%20%2B%20%5Cfrac%7Bp_A%20%2A%20%281%20-%20p_B%29%7D%7Bn_B%7D%7D)
![SE_{p_A-p_B} = \sqrt{\frac{0.23 * (1 - 0.23)}{100} + \frac{0.228* (1 - 0.228)}{250}}](https://tex.z-dn.net/?f=SE_%7Bp_A-p_B%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B0.23%20%2A%20%281%20-%200.23%29%7D%7B100%7D%20%2B%20%5Cfrac%7B0.228%2A%20%281%20-%200.228%29%7D%7B250%7D%7D)
![SE_{p_A-p_B} = \sqrt{\frac{0.1771}{100} + \frac{0.176016}{250}}](https://tex.z-dn.net/?f=SE_%7Bp_A-p_B%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B0.1771%7D%7B100%7D%20%2B%20%5Cfrac%7B0.176016%7D%7B250%7D%7D)
![SE_{p_A-p_B} = \sqrt{0.001771 + 0.000704064}](https://tex.z-dn.net/?f=SE_%7Bp_A-p_B%7D%20%3D%20%5Csqrt%7B0.001771%20%2B%200.000704064%7D)
![SE_{p_A-p_B} = \sqrt{0.002475064}](https://tex.z-dn.net/?f=SE_%7Bp_A-p_B%7D%20%3D%20%5Csqrt%7B0.002475064%7D)
![SE_{p_A-p_B} = 0.050](https://tex.z-dn.net/?f=SE_%7Bp_A-p_B%7D%20%3D%200.050)
Let's say hmm let's use the figures in thousands, so
model A costs 90 + 4 per year, so... ok.. after "t" years, then the cost for model A is then 90 + 4*t, or
A = 90 + 4t.
now, model B is 70 + 10 per year, so,... after "t" years, model B is then
B = 70 + 10 * t, or
B = 70 + 10t.
if we were to assume A < B, namely the cost of A is less than B, what would the years be? namely, what's "t"?
![\bf A\ \textless \ B\implies 90+4t \ \textless \ 70 + 10t\implies 90-70 \ \textless \ 10t-4t \\\\\\ 20 \ \textless \ 6t\implies \cfrac{20}{6}\ \textless \ t\implies \cfrac{10}{3}\ \textless \ t\implies 3\frac{1}{3}\ \textless \ t \\\\\\ \textit{3 years and }\frac{1}{3}\textit{ or \underline{3 years and 4 months}}](https://tex.z-dn.net/?f=%5Cbf%20A%5C%20%5Ctextless%20%5C%20B%5Cimplies%2090%2B4t%20%5C%20%5Ctextless%20%5C%20%2070%20%2B%2010t%5Cimplies%2090-70%20%5C%20%5Ctextless%20%5C%20%2010t-4t%0A%5C%5C%5C%5C%5C%5C%0A20%20%5C%20%5Ctextless%20%5C%20%206t%5Cimplies%20%5Ccfrac%7B20%7D%7B6%7D%5C%20%5Ctextless%20%5C%20t%5Cimplies%20%5Ccfrac%7B10%7D%7B3%7D%5C%20%5Ctextless%20%5C%20t%5Cimplies%203%5Cfrac%7B1%7D%7B3%7D%5C%20%5Ctextless%20%5C%20t%0A%5C%5C%5C%5C%5C%5C%0A%5Ctextit%7B3%20years%20and%20%7D%5Cfrac%7B1%7D%7B3%7D%5Ctextit%7B%20or%20%5Cunderline%7B3%20years%20and%204%20months%7D%7D)
after that period, then A becomes cheaper than B.
Answer:
b=60
Step-by-step explanation:
6+4/5b=9/10b
-4/5b -4/5b
6=1/10b
divide both sides by 1/10
b=60