Express the square root of -8 in i notation
1 answer:
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Answer:
Step-by-step explanation:
Hello!
The parameter of interest in this exercise is the population proportion of Asians that would welcome a person of other races in their family. Using the race of the welcomed one as categorizer we can define 3 variables:
X₁: Number of Asians that would welcome a white person into their families.
X₂: Number of Asians that would welcome a Latino person into their families.
X₃: Number of Asians that would welcome a black person into their families.
Now since we are working with the population that identifies as "Asians" the sample size will be: n= 251
Since the sample size is large enough (n≥30) you can apply the Central Limit Theorem and approximate the variable distribution to normal.
1. 95% CI for Asians that would welcome a white person.
If 79% would welcome a white person, then the expected value is:
E(X)= n*p= 251*0.79= 198.29
And the Standard deviation is:
V(X)= n*p*(1-p)= 251*0.79*0.21=41.6409
√V(X)= 6.45
You can construct the interval as:
E(X)±Z₁₋α/₂*√V(X)
198.29±1.965*6.45
[185.62;210.96]
With a 95% confidence level, you'd expect that the interval [185.62; 210.96] contains the number of Asian people that would welcome a White person in their family.
2. 95% CI for Asians that would welcome a Latino person.
If 71% would welcome a Latino person, then the expected value is:
E(X)= n*p= 251*0.71= 178.21
And the Standard deviation is:
V(X)= n*p*(1-p)= 251*0.71*0.29= 51.6809
√V(X)= 7.19
You can construct the interval as:
E(X)±Z₁₋α/₂*√V(X)
178.21±1.965*7.19
[164.08; 192.34]
With a 95% confidence level, you'd expect that the interval [164.08; 192.34] contains the number of Asian people that would welcome a Latino person in their family.
3. 95% CI for Asians that would welcome a Black person.
If 66% would welcome a Black person, then the expected value is:
E(X)= n*p= 251*0.66= 165.66
And the Standard deviation is:
V(X)= n*p*(1-p)= 251*0.66*0.34= 56.3244
√V(X)= 7.50
You can construct the interval as:
E(X)±Z₁₋α/₂*√V(X)
165.66±1.965*7.50
[150.92; 180.40]
With a 95% confidence level, you'd expect that the interval [150.92; 180.40] contains the number of Asian people that would welcome a Black person in their family.
I hope it helps!
Answer:
The 99% confidence interval estimate of the mean body temperature of all healthy humans is (97.955F, 98.245F).
98.6F is above the upper end of the interval, which means that the sample suggests that the mean body temperature could be lower than 98.6F.
Step-by-step explanation:
The first step is finding the confidence interval
The sample size is 103.
The first step to solve this problem is finding how many degrees of freedom there are, that is, the sample size subtracted by 1. So
Then, we need to subtract one by the confidence level and divide by 2. So:
Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 102 and 0.005 in the t-distribution table, we have .
Now, we need to find the standard deviation of the sample. That is:
Now, we multiply T and s
For the lower end of the interval, we subtract the mean by M. So 98.1 - 0.145 = 97.955F.
For the upper end of the interval, we add the mean to M. So 98.1 + 0.145 = 98.245F.
The 99% confidence interval estimate of the mean body temperature of all healthy humans is (97.955F, 98.245F).
98.6F is above the upper end of the interval, which means that the sample suggests that the mean body temperature could be lower than 98.6F.
Answer:
0.0000805
Step-by-step explanation:
= 0.0000805