Suppose the ages at retirement for workers in the U. S. are normally distributed with an unknown population mean and standard deviation. If a random sample of 43 retirees is taken to estimate the mean retirement age, what t-score should be used to find a 80% confidence interval estimate for the population mean?

1.302

Step-by-step explanation:

The sample used was 43 retirees, so n=43. To find the degrees of freedom:

df=n−1=43−1=42

The confidence level is given in the scenario: 80%. So,

α=1−CL=1−0.8=0.2

But we want to use the value for α2, which is 0.22=0.1. Using the table, we need to find the row for 42 degrees of freedom, and the column for t0.1. So, the t-score we would use to find the 80% confidence interval is 1.302.

We could also use a calculator and input 1−α2=0.9 and the degrees of freedom into invT, in which case we would enter invT(0.9,42).

the reporter's right

step-by-step explanation:

1/4 = .25 / 25%

30 miles

step-by-step explanation:

we know that 60 seconds equals 1 minute and 60 minutes 1 hour

44 ft       60 sec     60 min     158400   ft

  *   *   =

1 sec       1 min       1 hour         hour

we know that there are 5280 ft per mile

158400   ft       1 mile         30 miles

  *   =

1   hour               5280 ft     1 hour

so in 1 hour   it can travel 30 miles