
How do you express the confidence interval $0.111 < p < 0.333$ in the form of $p\pm E$?
Answer
539.1k+ views
Hint: For this problem we need to write the given confidence interval in form of $p\pm E$. We know that the $p$ denotes the sample proportion and $E$ denotes the margin of error. First, we will calculate the margin of error and after that we will use this value to calculate the sample proportion. We know that the margin error is the half of the difference between the given limits. So, we will calculate the margin error for the given limits. Now we will use this value and consider either lower limit or upper limit to calculate the sample proportion. If you consider the upper limit, we will subtract the margin of error from the upper limit. If you consider the lower limit, we will add the margin of error to the lower limit to calculate the sample proportion. After getting both the values we will write it in required form.
Complete step by step solution:
Given confidence interval $0.111 < p < 0.333$.
The upper limit in the above interval is $0.333$.
The lower limit in the above interval is $0.111$.
Difference between the both the limits is $0.333-0.111=0.222$.
Half of the difference between the both the limits is $\dfrac{0.222}{2}=0.111$.
Hence the Margin of error is $E=0.111$.
Considering the lower limit of the given interval which is $0.111$.
To calculate the sample proportion, we are going to add the margin of error to the lower limit then we will get
$p=0.111+0.111=0.222$
Now the confidence interval in $p\pm E$ from is $0.222\pm 0.111$.
Note:
We can also calculate the sample proportion without calculating the margin of error. We have the formula for the sample proportion which is
$p=\dfrac{U.L+L.L}{2}$
So, we will calculate the average of the given limits and that will be our sample proportion.
Complete step by step solution:
Given confidence interval $0.111 < p < 0.333$.
The upper limit in the above interval is $0.333$.
The lower limit in the above interval is $0.111$.
Difference between the both the limits is $0.333-0.111=0.222$.
Half of the difference between the both the limits is $\dfrac{0.222}{2}=0.111$.
Hence the Margin of error is $E=0.111$.
Considering the lower limit of the given interval which is $0.111$.
To calculate the sample proportion, we are going to add the margin of error to the lower limit then we will get
$p=0.111+0.111=0.222$
Now the confidence interval in $p\pm E$ from is $0.222\pm 0.111$.
Note:
We can also calculate the sample proportion without calculating the margin of error. We have the formula for the sample proportion which is
$p=\dfrac{U.L+L.L}{2}$
So, we will calculate the average of the given limits and that will be our sample proportion.
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