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Answer :
To determine whether the large counts condition is met when constructing a confidence interval for the proportion [tex]\( p \)[/tex], you need to check two values: [tex]\( n \hat{p} \)[/tex] and [tex]\( n(1 - \hat{p}) \)[/tex].
Here's how we calculate these two values:
1. Calculate [tex]\( n \hat{p} \)[/tex]:
Given [tex]\( n = 50 \)[/tex] and [tex]\( \hat{p} = 0.9 \)[/tex], we find:
[tex]\[
n \hat{p} = 50 \times 0.9 = 45
\][/tex]
2. Calculate [tex]\( n(1 - \hat{p}) \)[/tex]:
[tex]\[
n(1 - \hat{p}) = 50 \times (1 - 0.9) = 50 \times 0.1 = 5
\][/tex]
The large counts condition requires both [tex]\( n \hat{p} \)[/tex] and [tex]\( n(1 - \hat{p}) \)[/tex] to be at least 10. From our calculations:
- [tex]\( n \hat{p} = 45 \)[/tex], which is greater than 10.
- [tex]\( n(1 - \hat{p}) = 5 \)[/tex], which is less than 10.
Since [tex]\( n(1 - \hat{p}) \)[/tex] does not meet the requirement of being at least 10, the large counts condition is not met.
The correct answer to the question is:
No, [tex]$n \hat{p}$[/tex] and [tex]$n(1-\hat{p})$[/tex] are not both at least 10.
Here's how we calculate these two values:
1. Calculate [tex]\( n \hat{p} \)[/tex]:
Given [tex]\( n = 50 \)[/tex] and [tex]\( \hat{p} = 0.9 \)[/tex], we find:
[tex]\[
n \hat{p} = 50 \times 0.9 = 45
\][/tex]
2. Calculate [tex]\( n(1 - \hat{p}) \)[/tex]:
[tex]\[
n(1 - \hat{p}) = 50 \times (1 - 0.9) = 50 \times 0.1 = 5
\][/tex]
The large counts condition requires both [tex]\( n \hat{p} \)[/tex] and [tex]\( n(1 - \hat{p}) \)[/tex] to be at least 10. From our calculations:
- [tex]\( n \hat{p} = 45 \)[/tex], which is greater than 10.
- [tex]\( n(1 - \hat{p}) = 5 \)[/tex], which is less than 10.
Since [tex]\( n(1 - \hat{p}) \)[/tex] does not meet the requirement of being at least 10, the large counts condition is not met.
The correct answer to the question is:
No, [tex]$n \hat{p}$[/tex] and [tex]$n(1-\hat{p})$[/tex] are not both at least 10.
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