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How much money does the average professional football fan spend on food at a single football game? that question was posed to 10 randomly selected football fans. the sample results provided a sample mean and standard deviation of $14.00 and $2.50, respectively. use this information to construct a 90% confidence interval for the mean?

User Zavael
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Hi!

When the mean (
\bar{x}), the number of samples (n) and standard deviation (σ) are known, you can apply the following equation to determine confidence intervals. The value of z at 90% confidence can be found in tables (It is 1,645).


(\bar{x}-z* (\sigma)/( √(n) ),\bar{x}+z* (\sigma)/( √(n) )) \\ \\ (14-1,645* (2,50)/( √(10) ),14+1,645* (2,50)/( √(10) ) \\ \\ (12,70\$;15,30\$)

Have a nice day!
User Brent Eicher
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