How much money does the average professional football fan spend on food at a single football game? That question was posed to 60 randomly selected football fans. The sampled results show thatthe sample mean was $70.00 and prior sampling indicated that the population standard deviation was $17.50. Use this information to create a 95 percent confidence interval for the the average professional football fan spend on food at a single football game.

$70.00 +/- $4.43

= ( $65.57, $74.43)

Therefore at 95% confidence interval (a,b) = ( $65.57, $74.43)

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;

Sample Mean x = $70.00

Standard deviation r = $17.50

Number of samples n = 60

Confidence interval = 95%

z(at 95% confidence) = 1.96

Substituting the values we have;

$70.00+/-1.96($17.50/√60)

$70.00+/-1.96($2.259240285287)

$70.00+/-$4.428110959163

$70.00+/-$4.43

= ( $65.57, $74.43)

Therefore at 95% confidence interval (a,b) = ( $65.57, $74.43)

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step-by-step explanation:

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