| Title | Author | Created | Published | Tags | | ------------------------ | ---------- | ----------------- | ----------------- | -------------------------------------------------- | | Inferences in Population | Jon Marien | November 20, 2025 | November 20, 2025 | [[#classes\|#classes]], [[#MATH26367\|#MATH26367]] | # Inferences $ \hat{p}-z_{\alpha/2}\sqrt{\frac{\hat{p}\hat{q}}n}<p<\hat{p}+z_{\alpha/2}\sqrt{\frac{\hat{p}\hat{q}}n}$ $\hat{p}$ is the sample proportion -- that is, the proportion of a specific outcome or characteristic observed in a sample of data.​ $ \hat{p}=\frac xn$ $\hat{q}$ represents the sample proportion of failures — that is, the proportion of the sample that does not have the specific characteristic or outcome you are measuring. It is the complement of $\hat{p}$. $ \hat{q}=1-\hat{p}$ Error: $E=CI-\sqrt{\frac{\hat{p}\hat{q}}n}$ Minimum sample size: $n=\hat{p}\hat{q}\Bigg(\frac{z_{\alpha_2}}{E}\Bigg)^2$ If necessary, round up to obtain a whole number. This formula can be found by solving the margin of error value for $n$: $(2E=z_{\alpha/2}\sqrt{\frac{\hat{p}\hat{q}}n})$ $ \begin{array}{ccc}\hat{p}&\hat{q}&\hat{p}\hat{q}\\\\\hline0.1&0.9&0.09\\0.2&0.8&0.16\\0.3&0.7&0.21\\0.4&0.6&0.24\\0.5&0.5&0.25\\0.6&0.4&0.24\\0.7&0.3&0.21\\0.8&0.2&0.16\\0.9&0.1&0.09\end{array}$ # Example ![[image-1042.png]]Home Broadband Internet A researcher wants to estimate with 98% confidence the proportion of Canadians who have high-speed Internet connection at home. A previous study showed that 52% of Canadian homes had have high-speed Internet connection at home. The researcher wants to be accurate within 3% of the true proportion. Find the minimum sample size necessary. Solution: $\text{Since z}_{\alpha/2}=2.33,\text{E}=0.03,\:\hat{p}=0.52,\hat{q}=0.48$ $ n=\hat{p}\hat{q}\Bigg(\frac{z_{\alpha/2}}E\Bigg)^2=0.48(0.52)\Bigg(\frac{2.33}{0.03}\Bigg)^2=1505.6$ which, when rounded up, indicates that **1506 homes** need to be sampled.