Worked solution P67-Easy-1. The statistic is t=2.535 with df=24; p=0.0182. $$t=\frac{53.60-50}{7.10/\sqrt{25}}=2.535.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Use a random sample or randomized design, independence, and a distribution without problematic skew or outliers for the sample size. Error to reject: The t procedure accounts for estimating sigma with s; substituting a z distribution ignores that uncertainty.

Worked solution P67-Easy-2. The paired statistic is t=3.541 with df=20; p=0.0021. $$t=\frac{3.40}{4.40/\sqrt{21}}=3.541.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Differences must be independent across pairs and approximately normal or supported by adequate sample size without severe outliers. Error to reject: The analysis uses one sample of differences, not two independent samples.

Worked solution P67-Easy-3. The statistic is t=2.425 with df=26; p=0.0226. $$t=\frac{53.50-50}{7.50/\sqrt{27}}=2.425.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Use a random sample or randomized design, independence, and a distribution without problematic skew or outliers for the sample size. Error to reject: The t procedure accounts for estimating sigma with s; substituting a z distribution ignores that uncertainty.

Worked solution P67-Easy-4. The paired statistic is t=3.681 with df=22; p=0.0013. $$t=\frac{3.30}{4.30/\sqrt{23}}=3.681.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Differences must be independent across pairs and approximately normal or supported by adequate sample size without severe outliers. Error to reject: The analysis uses one sample of differences, not two independent samples.

Worked solution P67-Easy-5. The statistic is t=2.662 with df=28; p=0.0127. $$t=\frac{54.40-50}{8.90/\sqrt{29}}=2.662.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Use a random sample or randomized design, independence, and a distribution without problematic skew or outliers for the sample size. Error to reject: The t procedure accounts for estimating sigma with s; substituting a z distribution ignores that uncertainty.

Worked solution P67-Easy-6. The paired statistic is t=2.449 with df=24; p=0.0220. $$t=\frac{2.40}{4.90/\sqrt{25}}=2.449.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Differences must be independent across pairs and approximately normal or supported by adequate sample size without severe outliers. Error to reject: The analysis uses one sample of differences, not two independent samples.

Worked solution P67-Easy-7. The statistic is t=2.007 with df=30; p=0.0538. $$t=\frac{53.10-50}{8.60/\sqrt{31}}=2.007.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Use a random sample or randomized design, independence, and a distribution without problematic skew or outliers for the sample size. Error to reject: The t procedure accounts for estimating sigma with s; substituting a z distribution ignores that uncertainty.

Worked solution P67-Easy-8. The paired statistic is t=3.792 with df=26; p=0.0008. $$t=\frac{2.70}{3.70/\sqrt{27}}=3.792.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Differences must be independent across pairs and approximately normal or supported by adequate sample size without severe outliers. Error to reject: The analysis uses one sample of differences, not two independent samples.

Worked solution P67-Easy-9. The statistic is t=2.667 with df=32; p=0.0119. $$t=\frac{53.90-50}{8.40/\sqrt{33}}=2.667.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Use a random sample or randomized design, independence, and a distribution without problematic skew or outliers for the sample size. Error to reject: The t procedure accounts for estimating sigma with s; substituting a z distribution ignores that uncertainty.

Worked solution P67-Easy-10. The paired statistic is t=1.948 with df=28; p=0.0615. $$t=\frac{1.70}{4.70/\sqrt{29}}=1.948.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Differences must be independent across pairs and approximately normal or supported by adequate sample size without severe outliers. Error to reject: The analysis uses one sample of differences, not two independent samples.

Worked solution P67-Easy-11. The statistic is t=3.295 with df=34; p=0.0023. $$t=\frac{54.40-50}{7.90/\sqrt{35}}=3.295.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Use a random sample or randomized design, independence, and a distribution without problematic skew or outliers for the sample size. Error to reject: The t procedure accounts for estimating sigma with s; substituting a z distribution ignores that uncertainty.

Worked solution P67-Easy-12. The paired statistic is t=1.979 with df=20; p=0.0618. $$t=\frac{1.90}{4.40/\sqrt{21}}=1.979.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Differences must be independent across pairs and approximately normal or supported by adequate sample size without severe outliers. Error to reject: The analysis uses one sample of differences, not two independent samples.

Worked solution P67-Easy-13. The statistic is t=3.388 with df=36; p=0.0017. $$t=\frac{54.40-50}{7.90/\sqrt{37}}=3.388.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Use a random sample or randomized design, independence, and a distribution without problematic skew or outliers for the sample size. Error to reject: The t procedure accounts for estimating sigma with s; substituting a z distribution ignores that uncertainty.

Worked solution P67-Easy-14. The paired statistic is t=3.328 with df=22; p=0.0031. $$t=\frac{3.40}{4.90/\sqrt{23}}=3.328.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Differences must be independent across pairs and approximately normal or supported by adequate sample size without severe outliers. Error to reject: The analysis uses one sample of differences, not two independent samples.

Worked solution P67-Easy-15. The statistic is t=3.291 with df=38; p=0.0022. $$t=\frac{53.90-50}{7.40/\sqrt{39}}=3.291.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Use a random sample or randomized design, independence, and a distribution without problematic skew or outliers for the sample size. Error to reject: The t procedure accounts for estimating sigma with s; substituting a z distribution ignores that uncertainty.

Worked solution P67-Easy-16. The paired statistic is t=4.286 with df=24; p=0.0003. $$t=\frac{3.00}{3.50/\sqrt{25}}=4.286.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Differences must be independent across pairs and approximately normal or supported by adequate sample size without severe outliers. Error to reject: The analysis uses one sample of differences, not two independent samples.

Worked solution P67-Easy-17. The statistic is t=1.500 with df=25; p=0.1462. $$t=\frac{52.50-50}{8.50/\sqrt{26}}=1.500.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Use a random sample or randomized design, independence, and a distribution without problematic skew or outliers for the sample size. Error to reject: The t procedure accounts for estimating sigma with s; substituting a z distribution ignores that uncertainty.

Worked solution P67-Easy-18. The paired statistic is t=3.929 with df=26; p=0.0006. $$t=\frac{3.10}{4.10/\sqrt{27}}=3.929.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Differences must be independent across pairs and approximately normal or supported by adequate sample size without severe outliers. Error to reject: The analysis uses one sample of differences, not two independent samples.

Worked solution P67-Easy-19. The statistic is t=2.772 with df=27; p=0.0100. $$t=\frac{54.40-50}{8.40/\sqrt{28}}=2.772.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Use a random sample or randomized design, independence, and a distribution without problematic skew or outliers for the sample size. Error to reject: The t procedure accounts for estimating sigma with s; substituting a z distribution ignores that uncertainty.

Worked solution P67-Tough-1. The statistic is t=2.000 with df=24; p=0.0569. $$t=\frac{53.00-50}{7.50/\sqrt{25}}=2.000.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Use a random sample or randomized design, independence, and a distribution without problematic skew or outliers for the sample size. Error to reject: The t procedure accounts for estimating sigma with s; substituting a z distribution ignores that uncertainty.

Worked solution P67-Tough-2. The paired statistic is t=2.245 with df=20; p=0.0363. $$t=\frac{2.40}{4.90/\sqrt{21}}=2.245.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Differences must be independent across pairs and approximately normal or supported by adequate sample size without severe outliers. Error to reject: The analysis uses one sample of differences, not two independent samples.

Worked solution P67-Tough-3. The statistic is t=1.856 with df=26; p=0.0749. $$t=\frac{52.50-50}{7.00/\sqrt{27}}=1.856.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Use a random sample or randomized design, independence, and a distribution without problematic skew or outliers for the sample size. Error to reject: The t procedure accounts for estimating sigma with s; substituting a z distribution ignores that uncertainty.

Worked solution P67-Tough-4. The paired statistic is t=4.165 with df=22; p=0.0004. $$t=\frac{3.30}{3.80/\sqrt{23}}=4.165.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Differences must be independent across pairs and approximately normal or supported by adequate sample size without severe outliers. Error to reject: The analysis uses one sample of differences, not two independent samples.

Worked solution P67-Tough-5. The statistic is t=2.513 with df=28; p=0.0180. $$t=\frac{53.50-50}{7.50/\sqrt{29}}=2.513.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Use a random sample or randomized design, independence, and a distribution without problematic skew or outliers for the sample size. Error to reject: The t procedure accounts for estimating sigma with s; substituting a z distribution ignores that uncertainty.

Worked solution P67-Tough-6. The paired statistic is t=2.340 with df=24; p=0.0279. $$t=\frac{2.20}{4.70/\sqrt{25}}=2.340.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Differences must be independent across pairs and approximately normal or supported by adequate sample size without severe outliers. Error to reject: The analysis uses one sample of differences, not two independent samples.

Worked solution P67-Tough-7. The statistic is t=1.701 with df=30; p=0.0992. $$t=\frac{52.20-50}{7.20/\sqrt{31}}=1.701.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Use a random sample or randomized design, independence, and a distribution without problematic skew or outliers for the sample size. Error to reject: The t procedure accounts for estimating sigma with s; substituting a z distribution ignores that uncertainty.

Worked solution P67-Tough-8. The paired statistic is t=1.807 with df=26; p=0.0823. $$t=\frac{1.60}{4.60/\sqrt{27}}=1.807.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Differences must be independent across pairs and approximately normal or supported by adequate sample size without severe outliers. Error to reject: The analysis uses one sample of differences, not two independent samples.

Worked solution P67-Tough-9. The statistic is t=1.501 with df=32; p=0.1431. $$t=\frac{52.30-50}{8.80/\sqrt{33}}=1.501.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Use a random sample or randomized design, independence, and a distribution without problematic skew or outliers for the sample size. Error to reject: The t procedure accounts for estimating sigma with s; substituting a z distribution ignores that uncertainty.

Worked solution P67-Tough-10. The paired statistic is t=4.039 with df=28; p=0.0004. $$t=\frac{3.00}{4.00/\sqrt{29}}=4.039.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Differences must be independent across pairs and approximately normal or supported by adequate sample size without severe outliers. Error to reject: The analysis uses one sample of differences, not two independent samples.

Worked solution P67-Tough-11. The statistic is t=1.690 with df=34; p=0.1001. $$t=\frac{52.20-50}{7.70/\sqrt{35}}=1.690.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Use a random sample or randomized design, independence, and a distribution without problematic skew or outliers for the sample size. Error to reject: The t procedure accounts for estimating sigma with s; substituting a z distribution ignores that uncertainty.

Worked solution P67-Tough-12. The paired statistic is t=3.946 with df=20; p=0.0008. $$t=\frac{3.10}{3.60/\sqrt{21}}=3.946.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Differences must be independent across pairs and approximately normal or supported by adequate sample size without severe outliers. Error to reject: The analysis uses one sample of differences, not two independent samples.

Worked solution P67-Tough-13. The statistic is t=2.481 with df=36; p=0.0179. $$t=\frac{53.10-50}{7.60/\sqrt{37}}=2.481.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Use a random sample or randomized design, independence, and a distribution without problematic skew or outliers for the sample size. Error to reject: The t procedure accounts for estimating sigma with s; substituting a z distribution ignores that uncertainty.

Worked solution P67-Tough-14. The paired statistic is t=4.165 with df=22; p=0.0004. $$t=\frac{3.30}{3.80/\sqrt{23}}=4.165.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Differences must be independent across pairs and approximately normal or supported by adequate sample size without severe outliers. Error to reject: The analysis uses one sample of differences, not two independent samples.

Worked solution P67-Tough-15. The statistic is t=1.619 with df=38; p=0.1137. $$t=\frac{52.10-50}{8.10/\sqrt{39}}=1.619.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Use a random sample or randomized design, independence, and a distribution without problematic skew or outliers for the sample size. Error to reject: The t procedure accounts for estimating sigma with s; substituting a z distribution ignores that uncertainty.

Worked solution P67-Tough-16. The paired statistic is t=2.973 with df=24; p=0.0066. $$t=\frac{2.20}{3.70/\sqrt{25}}=2.973.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Differences must be independent across pairs and approximately normal or supported by adequate sample size without severe outliers. Error to reject: The analysis uses one sample of differences, not two independent samples.

Worked solution P67-Tough-17. The statistic is t=2.914 with df=25; p=0.0074. $$t=\frac{54.00-50}{7.00/\sqrt{26}}=2.914.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Use a random sample or randomized design, independence, and a distribution without problematic skew or outliers for the sample size. Error to reject: The t procedure accounts for estimating sigma with s; substituting a z distribution ignores that uncertainty.

Worked solution P67-Tough-18. The paired statistic is t=2.985 with df=26; p=0.0061. $$t=\frac{2.70}{4.70/\sqrt{27}}=2.985.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Differences must be independent across pairs and approximately normal or supported by adequate sample size without severe outliers. Error to reject: The analysis uses one sample of differences, not two independent samples.

Worked solution P67-Tough-19. The statistic is t=2.612 with df=27; p=0.0145. $$t=\frac{53.90-50}{7.90/\sqrt{28}}=2.612.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Use a random sample or randomized design, independence, and a distribution without problematic skew or outliers for the sample size. Error to reject: The t procedure accounts for estimating sigma with s; substituting a z distribution ignores that uncertainty.

Worked solution P67-Toughest-1. The statistic is t=2.403 with df=24; p=0.0244. $$t=\frac{53.70-50}{7.70/\sqrt{25}}=2.403.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Use a random sample or randomized design, independence, and a distribution without problematic skew or outliers for the sample size. Error to reject: The t procedure accounts for estimating sigma with s; substituting a z distribution ignores that uncertainty.

Worked solution P67-Toughest-2. The paired statistic is t=3.273 with df=20; p=0.0038. $$t=\frac{2.50}{3.50/\sqrt{21}}=3.273.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Differences must be independent across pairs and approximately normal or supported by adequate sample size without severe outliers. Error to reject: The analysis uses one sample of differences, not two independent samples.

Worked solution P67-Toughest-3. The statistic is t=1.386 with df=26; p=0.1776. $$t=\frac{52.00-50}{7.50/\sqrt{27}}=1.386.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Use a random sample or randomized design, independence, and a distribution without problematic skew or outliers for the sample size. Error to reject: The t procedure accounts for estimating sigma with s; substituting a z distribution ignores that uncertainty.

Worked solution P67-Toughest-4. The paired statistic is t=1.798 with df=22; p=0.0858. $$t=\frac{1.50}{4.00/\sqrt{23}}=1.798.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Differences must be independent across pairs and approximately normal or supported by adequate sample size without severe outliers. Error to reject: The analysis uses one sample of differences, not two independent samples.

Worked solution P67-Toughest-5. The statistic is t=1.488 with df=28; p=0.1479. $$t=\frac{52.10-50}{7.60/\sqrt{29}}=1.488.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Use a random sample or randomized design, independence, and a distribution without problematic skew or outliers for the sample size. Error to reject: The t procedure accounts for estimating sigma with s; substituting a z distribution ignores that uncertainty.

Worked solution P67-Toughest-6. The paired statistic is t=2.368 with df=24; p=0.0263. $$t=\frac{1.80}{3.80/\sqrt{25}}=2.368.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Differences must be independent across pairs and approximately normal or supported by adequate sample size without severe outliers. Error to reject: The analysis uses one sample of differences, not two independent samples.

Worked solution P67-Toughest-7. The statistic is t=1.683 with df=30; p=0.1027. $$t=\frac{52.60-50}{8.60/\sqrt{31}}=1.683.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Use a random sample or randomized design, independence, and a distribution without problematic skew or outliers for the sample size. Error to reject: The t procedure accounts for estimating sigma with s; substituting a z distribution ignores that uncertainty.

Worked solution P67-Toughest-8. The paired statistic is t=1.949 with df=26; p=0.0622. $$t=\frac{1.50}{4.00/\sqrt{27}}=1.949.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Differences must be independent across pairs and approximately normal or supported by adequate sample size without severe outliers. Error to reject: The analysis uses one sample of differences, not two independent samples.

Worked solution P67-Toughest-9. The statistic is t=1.783 with df=32; p=0.0841. $$t=\frac{52.70-50}{8.70/\sqrt{33}}=1.783.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Use a random sample or randomized design, independence, and a distribution without problematic skew or outliers for the sample size. Error to reject: The t procedure accounts for estimating sigma with s; substituting a z distribution ignores that uncertainty.

Worked solution P67-Toughest-10. The paired statistic is t=3.629 with df=28; p=0.0011. $$t=\frac{3.10}{4.60/\sqrt{29}}=3.629.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Differences must be independent across pairs and approximately normal or supported by adequate sample size without severe outliers. Error to reject: The analysis uses one sample of differences, not two independent samples.

Worked solution P67-Toughest-11. The statistic is t=3.485 with df=34; p=0.0014. $$t=\frac{54.30-50}{7.30/\sqrt{35}}=3.485.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Use a random sample or randomized design, independence, and a distribution without problematic skew or outliers for the sample size. Error to reject: The t procedure accounts for estimating sigma with s; substituting a z distribution ignores that uncertainty.

Worked solution P67-Toughest-12. The paired statistic is t=2.106 with df=20; p=0.0481. $$t=\frac{1.70}{3.70/\sqrt{21}}=2.106.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Differences must be independent across pairs and approximately normal or supported by adequate sample size without severe outliers. Error to reject: The analysis uses one sample of differences, not two independent samples.

Worked solution P67-Toughest-13. The statistic is t=1.632 with df=36; p=0.1114. $$t=\frac{52.20-50}{8.20/\sqrt{37}}=1.632.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Use a random sample or randomized design, independence, and a distribution without problematic skew or outliers for the sample size. Error to reject: The t procedure accounts for estimating sigma with s; substituting a z distribution ignores that uncertainty.

Worked solution P67-Toughest-14. The paired statistic is t=2.838 with df=22; p=0.0096. $$t=\frac{2.90}{4.90/\sqrt{23}}=2.838.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Differences must be independent across pairs and approximately normal or supported by adequate sample size without severe outliers. Error to reject: The analysis uses one sample of differences, not two independent samples.

Worked solution P67-Toughest-15. The statistic is t=2.082 with df=38; p=0.0442. $$t=\frac{52.50-50}{7.50/\sqrt{39}}=2.082.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Use a random sample or randomized design, independence, and a distribution without problematic skew or outliers for the sample size. Error to reject: The t procedure accounts for estimating sigma with s; substituting a z distribution ignores that uncertainty.

Worked solution P67-Toughest-16. The paired statistic is t=2.283 with df=24; p=0.0316. $$t=\frac{2.10}{4.60/\sqrt{25}}=2.283.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Differences must be independent across pairs and approximately normal or supported by adequate sample size without severe outliers. Error to reject: The analysis uses one sample of differences, not two independent samples.

Worked solution P67-Toughest-17. The statistic is t=2.234 with df=25; p=0.0346. $$t=\frac{53.90-50}{8.90/\sqrt{26}}=2.234.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Use a random sample or randomized design, independence, and a distribution without problematic skew or outliers for the sample size. Error to reject: The t procedure accounts for estimating sigma with s; substituting a z distribution ignores that uncertainty.

Worked solution P67-Toughest-18. The paired statistic is t=2.103 with df=26; p=0.0453. $$t=\frac{1.70}{4.20/\sqrt{27}}=2.103.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Differences must be independent across pairs and approximately normal or supported by adequate sample size without severe outliers. Error to reject: The analysis uses one sample of differences, not two independent samples.

Worked solution P67-Toughest-19. The statistic is t=1.724 with df=27; p=0.0961. $$t=\frac{52.90-50}{8.90/\sqrt{28}}=1.724.$$ Interpretation: The p-value quantifies evidence against the null mean; the design determines whether the result generalizes or supports causation. Validity: Use a random sample or randomized design, independence, and a distribution without problematic skew or outliers for the sample size. Error to reject: The t procedure accounts for estimating sigma with s; substituting a z distribution ignores that uncertainty.