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A sports marketing company interviewed 250 Supersport United fans and has found that the average fan spends R101 on non-ticket purchases at the stadium when he/she attends a home match. This includes money spent on food, drinks and merchandise. Assume that the standard deviation is R39.

Find the lower and upper limits for the average amount spent on non-ticket purchases per match by Supersport United fans. Use a confidence level of 90%. Round off your answers to 2 decimal places (4)​

Sagot :

raphealnwobi

The 90% confidence interval is between R96.94 and R105.06

The confidence (C) = 90% = 0.9, Mean (μ) = 101, standard deviation (σ) = 39, sample size (n) = 250

α = 1 - C = 1 - 0.9 = 0.1

α/2 = 0.1 / 2 = 0.05

The z score of α/2 corresponds with the z score of 0.45 (0.5 - 0.05) which is equal to 1.645

The margin of error (E) is given by:

[tex]E=Z_\frac{\alpha }{2} *\frac{\sigma}{\sqrt{n} } \\\\E=1.645 *\frac{39}{\sqrt{250} } =4.06[/tex]

The confidence interval = μ ± E = 101 ± 4.06 = (96.94, 105.06)

Therefore the 90% confidence interval is between R96.94 and R105.06

Find out more at: https://brainly.com/question/18914334

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