Lesson 9.2: Conditions and P-values for Significance Tests for a Proportion

Introduction to Significance Tests: Significance tests are employed to determine if experimental or observational results provide enough evidence to reject a null hypothesis ( $H_0$ ) in favor of an alternative hypothesis ( $H_a$ ). Previously, simulations were used to estimate P-values; however, standardized formulas allow for more precise calculations.
Example Case: Mrs. Gallas’s Free Throw Accuracy: - Mrs. Gallas claims an 80% free throw shooting percentage ( $p = 0.80$ ). - In a sample of $50$ free throws, she makes $32$ , resulting in a sample proportion ( $\hat{p}$ ) of $\frac{32}{50} = 0.64$ .

Parameter Definition: The parameter $p$ represents the true proportion of free throw makes for Mrs. Gallas.
The Null Hypothesis ( $H_0$ ): The claim being tested, assumed to be true until proven otherwise. - $H_0: p = 0.80$
The Alternative Hypothesis ( $H_a$ ): The claim we are looking for evidence for. - $H_a: p < 0.80$

Before calculating the test statistic and P-value, three conditions must be satisfied to ensure the validity of the inference:

1. Random Condition: The data must come from a random sample or a randomized experiment. - Significance: This allows us to generalize the results to the population of all free throws. - Status in Mrs. Gallas Case: Yes, the sample is assumed to be representative.
2. 10% Condition: When sampling without replacement, the sample size ( $n$ ) should be less than or equal to 10% of the population ( $N$ ). - Formula: $n \leq 0.10N$ - Application: $50 \leq 10\% \text{ (all free throws)}$ . This confirms that sampling without replacement is acceptable.
3. Large Counts Condition: Both the expected number of successes and failures must be at least $10$ . - Calculations: - $np = 50 \times 0.80 = 40 \geq 10$ - $n(1-p) = 50 \times (1 - 0.80) = 10 \geq 10$ - Significance: Meeting this condition ensures that the sampling distribution of $\hat{p}$ is approximately normal.

Mean ( $\mu_{\hat{p}}$ ): Equivalent to the hypothesized population proportion. - $\mu_{\hat{p}} = p = 0.80$
Standard Deviation ( $\sigma_{\hat{p}}$ ): Calculated using the formula for the standard deviation of a proportion. - $\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.80(0.20)}{50}} = 0.057$
Distribution Notation: The sampling distribution can be modeled as approximately normal: $N(0.80, 0.057)$ .
Normal Curve Labeling: The mean is centered at $0.80$ . Standard deviations (SD) are marked as follows: - $-1 \text{ SD}: 0.743$ - $-2 \text{ SD}: 0.686$ - $-3 \text{ SD}: 0.629$ - $+1 \text{ SD}: 0.857$ - $+2 \text{ SD}: 0.914$ - $+3 \text{ SD}: 0.971$

Test Statistic Formula: A measure of how far the sample statistic deviates from the null hypothesis parameter in units of standard deviation. - $z = \frac{\text{Statistic} - \text{Parameter}}{\text{Standard Deviation of Statistic}}$ - $z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$
Calculation for Mrs. Gallas ( $n=50$ , $\hat{p}=0.64$ ): - $z = \frac{0.64 - 0.80}{0.057} = -2.81$ - Interpretation: The observed value $\hat{p} = 0.64$ is $2.81$ standard deviations below the hypothesized mean ( $0.80$ ).
P-value Calculation: The probability of obtaining a sample result as extreme or more extreme than the one observed, assuming $H_0$ is true. - $P(\hat{p} \leq 0.64 \mid p = 0.80) = 0.002$
Conclusion Criteria: Generally, if the P-value is less than the significance level ( $\alpha = 0.05$ ), we reject $H_0$ . - Result: Because $0.002 < 0.05$ , we reject $H_0$ . We have convincing evidence that Mrs. Gallas is a less than 80% free throw shooter.

Hypothetical Case: Suppose Mrs. Gallas made $36/50$ shots. - Sample Proportion ( $\hat{p}$ ): $\frac{36}{50} = 0.72$ - Test Statistic ( $z$ ): $z = \frac{0.72 - 0.80}{0.057} = -1.40$ - P-value: $0.081$
Interpretation of P-value: Assuming $H_0$ is true ( $p = 0.80$ ), there is a $0.081$ probability of getting a sample proportion of $0.72$ or less purely by chance.
Decision: Because $0.081 > 0.05$ , we fail to reject $H_0$ .
Conclusion: We do not have convincing evidence that Mrs. Gallas is a less than 80% free throw shooter.

Conditions for Significance Test (LTA): - Random: Random sample or assignment. - 10%: $n \leq 0.10N$ . - Large Counts: $np \geq 10$ and $n(1-p) \geq 10$ .
Calculation (LTAZ): - $z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$
Interpretation of P-value: The probability of observing a result at least as extreme as the sample statistic by chance, given that the null hypothesis is true.

Context: Scientists believe $10\%$ of the population perceives a soapy flavor in cilantro. Ebise believes the proportion is higher among teenagers.
Study Details: Ebise takes a random sample of $n = 200$ teenagers and finds $25$ identify the soapy flavor.
1. Hypotheses: - $H_0: p = 0.10$ - $H_a: p > 0.10$ - Parameter definition ( $p$ ): The true proportion of all teenagers who get a soapy flavor for cilantro.
2. Evidence Assessment: - $\hat{p} = \frac{25}{200} = 0.125$ ( $12.5\%$ - This provides some evidence for $H_a$ because $12.5\% > 10\%$ .
3. Check Conditions: - Random: The problem states a "random sample of 200 teenagers," allowing generalization to the population. - 10%: $200 \leq 10\% \text{ (all teenagers)}$ , so sampling without replacement is acceptable. - Large Counts: - $200(0.10) = 20 \geq 10$ - $200(0.90) = 180 \geq 10$ - Conclusion: The sampling distribution of $\hat{p}$ is approximately normal.
4. Standardized Test Statistic and P-value: - $z = \frac{0.125 - 0.10}{\sqrt{\frac{0.10(0.90)}{200}}} = 1.18$ - P-value: $0.1193$
5. Interpretation of P-value: Assuming $H_0$ is true ( $p = 0.10$ ), there is a $0.1193$ probability of getting a sample proportion of $0.125$ or greater purely by chance.
6. Conclusion: - Compared against $\alpha = 0.05$ . - Because $0.1193 > 0.05$ , we fail to reject $H_0$ . We do not have convincing evidence that more than 10% of teenagers get a soapy flavor for cilantro.