Question

Question 16 1 pts A random sample of 56 fast food employees in a state yielded a mean hourly wage of $12.5. It is known that hourly wages in the state are normally distributed and the population standard deviation is $3.2. Build an 80% CI for the population mean hourly wage for fast food workers in the state. Keep three decimals in your Critical Value Zc, then round E to one decimal place before finding the endpoints of the interval. Enter your answer as a 6 digit string of numbers, where the first three digits are the lower endpoint without the decimal point and the last three digits are the upper endpoint without the decimal point. Example: 067134 would be entered for: 6.7 <ll< 13.4 Note: if you use the built-in Z-Interval command on a calculator you might be marked wrong. Follow my rounding instructions carefully to avoid grading errors.

Answer 1

80% Confidence Interval :-

X̅ ± Z( α /2) σ / √ ( n )

z critical value = 1.282 (From Z table)

margin of error E = Zc * $\sigma$ / sqrt (n)

= 1.282 * 3.2 / sqrt(56)

= 0.5

80% CI is

$\bar{x}$ - E < $\mu$ <  $\bar{x}$ + E

12.5 - 0.5 < $\mu$ < 12.5 + 0.5

12.0 < $\mu$ < 13.0

Answer = 120130