Justifying Claims Based on Confidence Interval - AP Statistics

1. Find the critical value for a 99% confidence interval using the standard normal distribution.

   2.576. For 99% confidence, $\alpha = 0.01$, so $Z_{0.005} = 2.576$.

2. Identify the lower bound of the confidence interval given $CI = (0.1, 0.2)$.

   Lower bound is 0.1. Lower bound is the first value in the interval.

3. State the formula for a confidence interval for a difference in population proportions.

   $(p_1 - p_2) \pm Z \times SE$. General form for any confidence level difference interval.

4. Calculate the upper bound of a confidence interval given $CI = (-0.05, 0.15)$.

   Upper bound is 0.15. Upper bound is the second value in the interval.

5. What does a confidence interval that does not include zero imply about a hypothesis test?

   Reject the null hypothesis. Excluding zero is equivalent to rejecting $H_0$.

6. What does a narrow confidence interval suggest about the difference between proportions?

   More precise estimate of the difference. Narrow intervals indicate less uncertainty in the estimate.

7. How does a higher confidence level affect the confidence interval width?

   Increases the width of the confidence interval. Higher confidence needs wider intervals for certainty.

8. Determine if zero is within the interval $(-0.03, 0.07)$ and interpret.

   Zero is within; no significant difference. Zero is between -0.03 and 0.07, so no significance.

9. What is the formula for the standard error of a difference in population proportions?

   $SE = \sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}$$. Formula combines variances of two independent sample proportions.

10. What is the null hypothesis for testing the difference between two population proportions?

    $H_0: p_1 - p_2 = 0$. Null hypothesis states no difference between proportions.

11. What is the role of the margin of error in a confidence interval?

    It defines the range of uncertainty around the point estimate. Margin of error quantifies the precision of the estimate.

12. Calculate the margin of error given $SE=0.05$ and $Z=1.96$.

    Margin of error is $1.96 \times 0.05 = 0.098$. Direct multiplication of critical value and standard error.

13. Identify the conclusion if the confidence interval completely above zero.

    $p_1$ is significantly greater than $p_2$. Entirely positive interval means $p_1 > p_2$ significantly.

14. Calculate the standard error with $p_1=0.7$, $p_2=0.6$, $n_1=200$, $n_2=180$.

    $SE = \sqrt{\frac{0.7 \times 0.3}{200} + \frac{0.6 \times 0.4}{180}}$. Substitute values into standard error formula for difference.

15. Find the point estimate for the difference in proportions given $p_1=0.5$, $p_2=0.3$.

    Point estimate is $0.5 - 0.3 = 0.2$. Point estimate is simply the difference $p_1 - p_2$.

16. What assumption must be met regarding sample independence in proportion tests?

    Samples must be independently and randomly selected. Independence ensures valid statistical inference.

17. How is the standard normal distribution used in constructing confidence intervals?

    To find the critical value $Z$. Standard normal provides critical values for intervals.

18. What is the effect of a smaller sample size on the standard error?

    Increases the standard error. Smaller samples have more variability and uncertainty.

19. What is the alternate hypothesis for a two-tailed test of proportion differences?

    $H_a: p_1 - p_2 \neq 0$. Two-tailed test checks for any difference from zero.

20. What does it mean if a 95% confidence interval for $p_1 - p_2$ does not include zero?

    There is a significant difference between $p_1$ and $p_2$. Zero excluded means difference is statistically significant.

21. What is indicated by a confidence interval that spans both negative and positive values?

    No significant difference between proportions. Including zero means no significant difference detected.

22. What does a confidence interval that includes zero indicate about the difference in proportions?

    It suggests no significant difference between proportions. Zero in the interval means the difference could be zero.

23. What significance level is typically used in constructing a 95% confidence interval?

    5% significance level (0.05). 95% confidence corresponds to $\alpha = 0.05$ significance.

24. What is the significance of the confidence level in interval estimation?

    Probability that the interval contains the true parameter. Confidence level shows probability of capturing true parameter.

25. What influences the choice of a confidence level in interval estimation?

    Desired precision and risk tolerance. Higher confidence requires more precision and certainty.

26. Which condition checks that each sample size is large enough for a proportion test?

    Both $np$ and $n(1-p)$ must be greater than 10. Ensures normal approximation validity for each sample.

27. What is the purpose of constructing a confidence interval for a difference in proportions?

    To estimate the range of the difference between two population proportions. Provides plausible range for the true difference.

28. Find the confidence interval lower bound given $CI = (0.12, 0.22)$.

    Lower bound is 0.12. Lower bound is the first value in the interval.

29. Identify the critical value for a 95% confidence interval using the standard normal distribution.

    1.96. For 95% confidence, $\alpha = 0.05$, so $Z_{\alpha/2} = 1.96$.

30. What happens to the confidence interval if the sample size is doubled?

    The width of the confidence interval decreases. Doubling sample size reduces standard error by $\sqrt{2}$.

31. What is the impact of increasing sample size on the confidence interval width?

    Decreases the width of the confidence interval. Larger samples reduce standard error, narrowing intervals.

32. Identify the interpretation of a confidence interval containing only positive values.

    $p_1$ is significantly greater than $p_2$. All positive values mean $p_1 > p_2$ significantly.

33. Choose the correct formula for calculating the pooled sample proportion.

    $\hat{p} = \frac{x_1 + x_2}{n_1 + n_2}$. Combines successes from both samples over total size.

34. Calculate the standard error for $p_1=0.45$, $p_2=0.35$, $n_1=120$, $n_2=130$.

    $SE = \sqrt{\frac{0.45 \times 0.55}{120} + \frac{0.35 \times 0.65}{130}}$. Substitute given values into the standard error formula.

35. Why is the standard normal distribution used in confidence interval calculations?

    To determine the critical value $Z$. Normal distribution provides appropriate critical values.

Study these flashcards