Worked solution P64-Easy-1. The statistic is z=2.952; the two-sided p-value is 0.0032. $$z=\frac{0.6320-0.50}{\sqrt{0.50(0.50)/125}}=2.952.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Easy-2. The statistic is z=1.425; the two-sided p-value is 0.1540. $$z=\frac{0.5635-0.50}{\sqrt{0.50(0.50)/126}}=1.425.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Easy-3. The statistic is z=1.331; the two-sided p-value is 0.1832. $$z=\frac{0.5591-0.50}{\sqrt{0.50(0.50)/127}}=1.331.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Easy-4. The statistic is z=2.121; the two-sided p-value is 0.0339. $$z=\frac{0.5938-0.50}{\sqrt{0.50(0.50)/128}}=2.121.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Easy-5. The statistic is z=2.553; the two-sided p-value is 0.0107. $$z=\frac{0.6124-0.50}{\sqrt{0.50(0.50)/129}}=2.553.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Easy-6. The statistic is z=2.982; the two-sided p-value is 0.0029. $$z=\frac{0.6308-0.50}{\sqrt{0.50(0.50)/130}}=2.982.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Easy-7. The statistic is z=1.311; the two-sided p-value is 0.1900. $$z=\frac{0.5573-0.50}{\sqrt{0.50(0.50)/131}}=1.311.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Easy-8. The statistic is z=2.959; the two-sided p-value is 0.0031. $$z=\frac{0.6288-0.50}{\sqrt{0.50(0.50)/132}}=2.959.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Easy-9. The statistic is z=1.648; the two-sided p-value is 0.0995. $$z=\frac{0.5714-0.50}{\sqrt{0.50(0.50)/133}}=1.648.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Easy-10. The statistic is z=3.455; the two-sided p-value is 0.0005. $$z=\frac{0.6493-0.50}{\sqrt{0.50(0.50)/134}}=3.455.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Easy-11. The statistic is z=3.184; the two-sided p-value is 0.0015. $$z=\frac{0.6370-0.50}{\sqrt{0.50(0.50)/135}}=3.184.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Easy-12. The statistic is z=2.058; the two-sided p-value is 0.0396. $$z=\frac{0.5882-0.50}{\sqrt{0.50(0.50)/136}}=2.058.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Easy-13. The statistic is z=2.990; the two-sided p-value is 0.0028. $$z=\frac{0.6277-0.50}{\sqrt{0.50(0.50)/137}}=2.990.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Easy-14. The statistic is z=1.703; the two-sided p-value is 0.0887. $$z=\frac{0.5725-0.50}{\sqrt{0.50(0.50)/138}}=1.703.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Easy-15. The statistic is z=3.478; the two-sided p-value is 0.0005. $$z=\frac{0.6475-0.50}{\sqrt{0.50(0.50)/139}}=3.478.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Easy-16. The statistic is z=2.874; the two-sided p-value is 0.0041. $$z=\frac{0.6214-0.50}{\sqrt{0.50(0.50)/140}}=2.874.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Easy-17. The statistic is z=1.937; the two-sided p-value is 0.0528. $$z=\frac{0.5816-0.50}{\sqrt{0.50(0.50)/141}}=1.937.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Easy-18. The statistic is z=3.525; the two-sided p-value is 0.0004. $$z=\frac{0.6479-0.50}{\sqrt{0.50(0.50)/142}}=3.525.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Easy-19. The statistic is z=3.094; the two-sided p-value is 0.0020. $$z=\frac{0.6294-0.50}{\sqrt{0.50(0.50)/143}}=3.094.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Tough-1. The statistic is z=1.342; the two-sided p-value is 0.1797. $$z=\frac{0.5600-0.50}{\sqrt{0.50(0.50)/125}}=1.342.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Tough-2. The statistic is z=1.960; the two-sided p-value is 0.0500. $$z=\frac{0.5873-0.50}{\sqrt{0.50(0.50)/126}}=1.960.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Tough-3. The statistic is z=3.106; the two-sided p-value is 0.0019. $$z=\frac{0.6378-0.50}{\sqrt{0.50(0.50)/127}}=3.106.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Tough-4. The statistic is z=1.414; the two-sided p-value is 0.1573. $$z=\frac{0.5625-0.50}{\sqrt{0.50(0.50)/128}}=1.414.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Tough-5. The statistic is z=2.201; the two-sided p-value is 0.0277. $$z=\frac{0.5969-0.50}{\sqrt{0.50(0.50)/129}}=2.201.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Tough-6. The statistic is z=2.807; the two-sided p-value is 0.0050. $$z=\frac{0.6231-0.50}{\sqrt{0.50(0.50)/130}}=2.807.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Tough-7. The statistic is z=1.311; the two-sided p-value is 0.1900. $$z=\frac{0.5573-0.50}{\sqrt{0.50(0.50)/131}}=1.311.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Tough-8. The statistic is z=2.785; the two-sided p-value is 0.0053. $$z=\frac{0.6212-0.50}{\sqrt{0.50(0.50)/132}}=2.785.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Tough-9. The statistic is z=3.208; the two-sided p-value is 0.0013. $$z=\frac{0.6391-0.50}{\sqrt{0.50(0.50)/133}}=3.208.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Tough-10. The statistic is z=1.555; the two-sided p-value is 0.1200. $$z=\frac{0.5672-0.50}{\sqrt{0.50(0.50)/134}}=1.555.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Tough-11. The statistic is z=1.463; the two-sided p-value is 0.1434. $$z=\frac{0.5630-0.50}{\sqrt{0.50(0.50)/135}}=1.463.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Tough-12. The statistic is z=2.058; the two-sided p-value is 0.0396. $$z=\frac{0.5882-0.50}{\sqrt{0.50(0.50)/136}}=2.058.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Tough-13. The statistic is z=2.136; the two-sided p-value is 0.0327. $$z=\frac{0.5912-0.50}{\sqrt{0.50(0.50)/137}}=2.136.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Tough-14. The statistic is z=2.043; the two-sided p-value is 0.0411. $$z=\frac{0.5870-0.50}{\sqrt{0.50(0.50)/138}}=2.043.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Tough-15. The statistic is z=1.951; the two-sided p-value is 0.0511. $$z=\frac{0.5827-0.50}{\sqrt{0.50(0.50)/139}}=1.951.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Tough-16. The statistic is z=3.043; the two-sided p-value is 0.0023. $$z=\frac{0.6286-0.50}{\sqrt{0.50(0.50)/140}}=3.043.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Tough-17. The statistic is z=3.284; the two-sided p-value is 0.0010. $$z=\frac{0.6383-0.50}{\sqrt{0.50(0.50)/141}}=3.284.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Tough-18. The statistic is z=2.182; the two-sided p-value is 0.0291. $$z=\frac{0.5915-0.50}{\sqrt{0.50(0.50)/142}}=2.182.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Tough-19. The statistic is z=2.425; the two-sided p-value is 0.0153. $$z=\frac{0.6014-0.50}{\sqrt{0.50(0.50)/143}}=2.425.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Toughest-1. The statistic is z=2.415; the two-sided p-value is 0.0157. $$z=\frac{0.6080-0.50}{\sqrt{0.50(0.50)/125}}=2.415.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Toughest-2. The statistic is z=1.604; the two-sided p-value is 0.1088. $$z=\frac{0.5714-0.50}{\sqrt{0.50(0.50)/126}}=1.604.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Toughest-3. The statistic is z=2.041; the two-sided p-value is 0.0413. $$z=\frac{0.5906-0.50}{\sqrt{0.50(0.50)/127}}=2.041.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Toughest-4. The statistic is z=3.359; the two-sided p-value is 0.0008. $$z=\frac{0.6484-0.50}{\sqrt{0.50(0.50)/128}}=3.359.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Toughest-5. The statistic is z=1.849; the two-sided p-value is 0.0645. $$z=\frac{0.5814-0.50}{\sqrt{0.50(0.50)/129}}=1.849.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Toughest-6. The statistic is z=1.579; the two-sided p-value is 0.1144. $$z=\frac{0.5692-0.50}{\sqrt{0.50(0.50)/130}}=1.579.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Toughest-7. The statistic is z=1.311; the two-sided p-value is 0.1900. $$z=\frac{0.5573-0.50}{\sqrt{0.50(0.50)/131}}=1.311.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Toughest-8. The statistic is z=2.611; the two-sided p-value is 0.0090. $$z=\frac{0.6136-0.50}{\sqrt{0.50(0.50)/132}}=2.611.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Toughest-9. The statistic is z=1.301; the two-sided p-value is 0.1934. $$z=\frac{0.5564-0.50}{\sqrt{0.50(0.50)/133}}=1.301.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Toughest-10. The statistic is z=1.555; the two-sided p-value is 0.1200. $$z=\frac{0.5672-0.50}{\sqrt{0.50(0.50)/134}}=1.555.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Toughest-11. The statistic is z=3.529; the two-sided p-value is 0.0004. $$z=\frac{0.6519-0.50}{\sqrt{0.50(0.50)/135}}=3.529.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Toughest-12. The statistic is z=1.372; the two-sided p-value is 0.1701. $$z=\frac{0.5588-0.50}{\sqrt{0.50(0.50)/136}}=1.372.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Toughest-13. The statistic is z=1.623; the two-sided p-value is 0.1045. $$z=\frac{0.5693-0.50}{\sqrt{0.50(0.50)/137}}=1.623.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Toughest-14. The statistic is z=3.065; the two-sided p-value is 0.0022. $$z=\frac{0.6304-0.50}{\sqrt{0.50(0.50)/138}}=3.065.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Toughest-15. The statistic is z=3.138; the two-sided p-value is 0.0017. $$z=\frac{0.6331-0.50}{\sqrt{0.50(0.50)/139}}=3.138.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Toughest-16. The statistic is z=3.381; the two-sided p-value is 0.0007. $$z=\frac{0.6429-0.50}{\sqrt{0.50(0.50)/140}}=3.381.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Toughest-17. The statistic is z=2.105; the two-sided p-value is 0.0353. $$z=\frac{0.5887-0.50}{\sqrt{0.50(0.50)/141}}=2.105.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Toughest-18. The statistic is z=3.021; the two-sided p-value is 0.0025. $$z=\frac{0.6268-0.50}{\sqrt{0.50(0.50)/142}}=3.021.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.

Worked solution P64-Toughest-19. The statistic is z=2.091; the two-sided p-value is 0.0366. $$z=\frac{0.5874-0.50}{\sqrt{0.50(0.50)/143}}=2.091.$$ Interpretation: Compare the p-value with the stated alpha and make a population conclusion in the direction specified by Ha. Validity: Check randomization, the 10 percent condition if relevant, and large counts using the null value p0. Error to reject: Use p0 in the test standard error; using hat p would turn the calculation into the interval standard error.