In a recent poll, 600 people were asked if they liked soccer, and 72% said they did. Based on this, construct a 99% confidence interval for the true population proportion of people who like soccer.As in the reading, in your calculations:--Use z = 1.645 for a 90% confidence interval--Use z = 2 for a 95% confidence interval--Use z = 2.576 for a 99% confidence interval.Give your answers as decimals, to 4 decimal places.Out of 500 people sampled, 80 had kids. Based on this, construct a 90% confidence interval for the true population proportion of people with kids.As in the reading, in your calculations:--Use z = 1.645 for a 90% confidence interval--Use z = 2 for a 95% confidence interval--Use z = 2.576 for a 99% confidence interval.Give your answers to three decimalsGive your answers as decimals, to three decimal places.

Answer:(0.6728, 0.7672)(0.1330, 0.1870)Step-by-step explanation:Given that in a recent poll, 600 people were asked if they liked soccer, and 72% said they did.Std error = [tex]\sqrt{\frac{pq}{n} } \\=\sqrt{\frac{0.72(0.28)}{600} } \\=0.0183[/tex]Margin of error for 99% we would use the value [tex]z = 2.576[/tex]Margin of error = [tex]2.576*SE\\=2.576*0.0183\\=0.0472[/tex]Confidence interval lower bound = [tex]0.72-0.0472=0.6728[/tex]Upper bound = [tex]0.72+0.0472=0.7672[/tex]99% confidence interval for the true population proportion of people who like soccer.=(0.6728, 0.7672)b) n =500Sample proportion p=[tex]\frac{80}{500} =0.16[/tex]Margin of error for 90% = [tex]1.645*\sqrt{\frac{0.16*0.84}{500} } \\\\=0.0270[/tex]90% confidence interval for the true population proportion of people with kids. =[tex](0.16-0.0270, 0.16+0.0270)\\=(0.1330, 0.1870)[/tex]