Question

As part of a comprehensive survey on personal finances, a random sample of 100 consumers was asked how much they spent using a debit card in 2017. The average for these 100 consumers is $7,790. The population standard deviation, LaTeX: \sigma ? , is known to be $500. What is the Lower Confidence Limit for the 99% confidence interval for the population mean amount spent using a debit card? Round your answer to the nearest dollar.

Answer 1

For 99% CI z value is 2.58 as

For 99% CI z value is 2.58 as P(- 2.58 < z < 2.58) = 0.99 Here we have use z distribution as sample size is large enough and also population deviation is known So Margin of Error = E = z * theta/squareroot n = 2.58 * 500/10 = 129 so Cl is mu plusminus 129 = (7661,7919) So Lower Confidence Limit is 7661

Here we have use z distribution as sample size is large enough and also population deviation is known

So MArgin of Error=

So CI is $\mu \pm E=7790 \pm 129=(7661,7919)$

So Lower Confidence Limit is 7661.