A random sample of 22 lunch orders at Noodles & Company showed a mean bill of $10.26 with a standard deviation of $5.21. Find the 99 percent confidence interval for the mean bill of
all lunch orders.

(7.115 ; 13.405)

Explanation:

Given :

Sample size, n = 22

Mean bill, μ = 10.26

Standard deviation, s = 5.21

To obtain the 99% confidence interval for the mean bill of all orders ;

Mean ± margin of error

Margin of Error = Tcritical * s/√n

Tcritical at 99%, df = n-1, 22 - 1 = 21

Tcritical = 2.831

Margin of Error = 2.831 * (5.21/√22) = 3.145

Confidence interval = 10.26 ± 3.145

Lower boundary = 10.26 - 3.145 = 7.115

Upper boundary = 10.26 + 3.145 = 13.405

Confidence interval :

(7.115 ; 13.405)

depends on what you mean by "group." for order, it would be coleoptera (beetles). for class, it would be insecta (insects). for phylum, it would be the arthropoda (animals with jointed exoskeletons).

explanation:

there is no correct list among the choices you've provided.

the list might look something like this:

-- baby human

-- piccolo

-- flute

-- soprano singer

-- trumpet

-- tenor singer

-- trombone

-- baritone singer

-- baritone horn

-- bass singer

-- tuba

since i had to make up the list myself, i made it easy on myself, and i didn't try to shuffle the string instruments and woodwinds into it too.