Question

1.) An online site presented this question, "Would the recent norovirus outbreak deter you from taking a cruise?" Among the 34,705 people who responded, 69% answered "yes." Use the sample data to construct a 90% confidence interval estimate for the proportion of the population of all people who would respond "yes" to that question. Does the confidence interval provide a good estimate of the population proportion?

______<p<______

Answer #1

sample size n= | 34705 | |

sample proportion p̂ =x/n= | 0.6900 | |

std error se= √(p*(1-p)/n) = | 0.0025 | |

for 90 % CI value of z= | 1.645 | |

margin of error E=z*std error = | 0.0041 | |

lower bound=p̂ -E = | 0.6859 | |

Upper bound=p̂ +E = | 0.6941 |

from above 90% confidence interval for population
proportion =(0.6856 <p< 0.6941) |

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