Topcis
- Comparing Two Proportions
- The sampling distribution of \hat{p}_1-\hat{p}_2 has the following properties:
- Shape: Approximately normal if the sample are large enough that large counts is met for both samples
- Center: Mean at p_1-p_2
- Spread: As long as each sample is no more than 10% of its population, the standard deviation is \sigma_{p_d}=\sqrt{\frac{\hat{p}_{1}(1-\hat{p}_{1})}{n_1}+\frac{\hat{p}_{2}(1-\hat{p}_{2})}{n_2}}
- If the conditions for estimating or testing a claim for a one sample proportion are met for each distribution, a test can be preformed with the difference.
- A two-sample z interval for p_1-p_2 is performed by the formula (\hat{p}_1-\hat{p}_2) \pm z^{*}\sqrt{\frac{\hat{p}_{1}(1-\hat{p}_{1})}{n_1}+\frac{\hat{p}_{2}(1-\hat{p}_{2})}{n_2}}
- A two-sample z test for p_1-p_2 is performed by first finding \hat{p}_C and then z=\frac{(\hat{p}_1-\hat{p}_2)-0}{\sqrt{\frac{\hat{p}_{C}(1-\hat{p}_{C})}{n_1}+\frac{\hat{p}_{C}(1-\hat{p}_{C})}{n_2}}}
- The sampling distribution of \hat{p}_1-\hat{p}_2 has the following properties:
- Comparing Two Means
- The sampling distribution of \bar{x}_1-\bar{x}_2 has the following properties:
- Shape: Normal if both population distributions are normal, otherwise approximately normal if both samples are large enough according to the CLT
- Center: Mean at \mu_1-\mu_2
- Spread: As long as each sample is no more then 10% of its population, the standard deviation is \sqrt{\frac{\sigma^{2}_1}{n_1}+\frac{\sigma^{2}_2}{n_2}}
- If the conditions for estimating or testing a claim for a one sample mean are ment for each distribution, a test can be preformed with the difference
- Since the population standard deviation is often never known, a t distribution should be used. To avoid using a long formula, df should equal the smallest of n_1-1 and n_2-1
- The sampling distribution of \bar{x}_1-\bar{x}_2 has the following properties:
Formulas
\LARGE
\sigma_{p_d}=\sqrt{\frac{\hat{p}_{1}(1-\hat{p}_{1})}{n_1}+\frac{\hat{p}_{2}(1-\hat{p}_{2})}{n_2}}
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\LARGE
(\hat{p}_1-\hat{p}_2) \pm z^{*}\sqrt{\frac{\hat{p}_{1}(1-\hat{p}_{1})}{n_1}+\frac{\hat{p}_{2}(1-\hat{p}_{2})}{n_2}}
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\LARGE
\hat{p}_C=\frac{\textrm{count of successes in both samples}}{\textrm{count of individuals in both samples}}=\frac{X_1+X_2}{n_1+n_2}
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\LARGE
z=\frac{(\hat{p}_1-\hat{p}_2)-0}{\sqrt{\frac{\hat{p}_{C}(1-\hat{p}_{C})}{n_1}+\frac{\hat{p}_{C}(1-\hat{p}_{C})}{n_2}}}
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\LARGE
\sigma_{\mu_d}=\sqrt{\frac{\sigma^{2}_1}{n_1}+\frac{\sigma^{2}_2}{n_2}}
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\LARGE
(\bar{x}_1-\bar{x}_2) \pm t^{*}\sqrt{\frac{\sigma^{2}_1}{n_1}+\frac{\sigma^{2}_2}{n_2}}
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\LARGE
t=\frac{(\bar{x}_1-\bar{x}_2)-(\mu_1-\mu_2)}{\sqrt{\frac{\sigma^{2}_1}{n_1}+\frac{\sigma^{2}_2}{n_2}}}
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