Question

1.) A back-of-the-envelope calculation for a 95% CI for the

population mean mu is

[xbar - 2*SE , xbar + 2*SE]

where xbar is the sample mean and SE is the standard error

of the sample mean. When xbar = 3.1 and SE 1.5,

what is the left end point of a back-of-the-envelope 95%

confidence interval for the population mean?

___________________________________________

2.) Suppose that 32 of the 1500 people that received an email we sent, clicked on the offer image. What is the estimate for the left end point of a 95% confidence interval for the proportion of people in the population that will click on an offer that we send them? (Round to 3 digits after the decimal point)

Answer #1

1)left end point of a back-of-the-envelope 95% confidence
interval for the population mean =3.1-2*1.5
**=0.1**

2)

sample success x = | 32 | |

sample size n= | 1500 | |

sample proportion p̂ =x/n= | 0.0213 | |

std error se= √(p*(1-p)/n) = | 0.0037 |

estimate for the left end point of a 95% confidence interval for
the proportion of people in the population that will click on an
offer that we send them =0.0213-2*0.0037
**=0.014**

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