To determine whether a normal distribution can be used to calculate the margin of error for a given sample size [tex]\( n \)[/tex] and sample proportion [tex]\( \hat{p} \)[/tex], we need to check two conditions:
1. [tex]\( n\hat{p} \geq 5 \)[/tex]
2. [tex]\( n(1 - \hat{p}) \geq 5 \)[/tex]
We will verify these conditions for each given pair of [tex]\( n \)[/tex] and [tex]\( \hat{p} \)[/tex]:
1. For [tex]\( n = 50 \)[/tex] and [tex]\( \hat{p} = 0.222 \)[/tex]:
- [tex]\( n\hat{p} = 50 \times 0.222 = 11.1 \)[/tex]
- [tex]\( n(1 - \hat{p}) = 50 \times (1 - 0.222) = 50 \times 0.778 = 38.9 \)[/tex]
Both conditions ([tex]\( 11.1 \geq 5 \)[/tex] and [tex]\( 38.9 \geq 5 \)[/tex]) are satisfied.
2. For [tex]\( n = 40 \)[/tex] and [tex]\( \hat{p} = 0.207 \)[/tex]:
- [tex]\( n\hat{p} = 40 \times 0.207 = 8.28 \)[/tex]
- [tex]\( n(1 - \hat{p}) = 40 \times (1 - 0.207) = 40 \times 0.793 = 31.72 \)[/tex]
Both conditions ([tex]\( 8.28 \geq 5 \)[/tex] and [tex]\( 31.72 \geq 5 \)[/tex]) are satisfied.
3. For [tex]\( n = 30 \)[/tex] and [tex]\( \hat{p} = 0.327 \)[/tex]:
- [tex]\( n\hat{p} = 30 \times 0.327 = 9.81 \)[/tex]
- [tex]\( n(1 - \hat{p}) = 30 \times (1 - 0.327) = 30 \times 0.673 = 20.19 \)[/tex]
Both conditions ([tex]\( 9.81 \geq 5 \)[/tex] and [tex]\( 20.19 \geq 5 \)[/tex]) are satisfied.
4. For [tex]\( n = 100 \)[/tex] and [tex]\( \hat{p} = 0.095 \)[/tex]:
- [tex]\( n\hat{p} = 100 \times 0.095 = 9.5 \)[/tex]
- [tex]\( n(1 - \hat{p}) = 100 \times (1 - 0.095) = 100 \times 0.905 = 90.5 \)[/tex]
Both conditions ([tex]\( 9.5 \geq 5 \)[/tex] and [tex]\( 90.5 \geq 5 \)[/tex]) are satisfied.
Since all the conditions are satisfied for all the given pairs, a normal distribution can be used to calculate the margin of error for all these sample sizes and proportions:
- [tex]\( n = 50 \)[/tex] and [tex]\( \hat{p} = 0.222 \)[/tex]
- [tex]\( n = 40 \)[/tex] and [tex]\( \hat{p} = 0.207 \)[/tex]
- [tex]\( n = 30 \)[/tex] and [tex]\( \hat{p} = 0.327 \)[/tex]
- [tex]\( n = 100 \)[/tex] and [tex]\( \hat{p} = 0.095 \)[/tex]
