In a sample of 22 people, the average cost of a cup of coffee is $2.70. Assume the population standard deviation is $0.93. What is the 90% confidence interval for the cost of a cup of coffee

$2.70+/-$0.33

= ( $2.37, $3.03)

Therefore, the 90% confidence interval (a,b) = ( $2.37, $3.03)

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x+/-zr/√n

Given that;

Mean x = $2.70

Standard deviation r = $0.93

Number of samples n = 22

Confidence interval = 90%

z(at 90% confidence) = 1.645

Substituting the values we have;

$2.70+/-1.645($0.93/√22)

$2.70+/-1.645($0.198276666210)

$2.70+/-$0.326165115916

$2.70+/-$0.33

= ( $2.37, $3.03)

Therefore, the 90% confidence interval (a,b) = ( $2.37, $3.03)

the circle one

1gm = 1000mg so 0.1gm = 1000mg

first, convert the weight to kg

160lb * 1kg/2.2lb = 72.7kg

next, the dose

72.7kg * 100mg/kg = 7270mg