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13 December, 15:18

A survey of several 9 to 11 year olds recorded the following amounts spent on a trip to the mall: $10.31,$17.22,$26.62,$22.84 Construct the 98% confidence interval for the average amount spent by 9 to 11 year olds on a trip to the mall. Assume the population is approximately normal. Step 1 of 4 : Calculate the sample mean for the given sample data. Round your answer to two decimal places.

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  1. 13 December, 18:59
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    Step-by-step explanation:

    From the information given,

    Mean, μ = (10.31 + 17.22 + 26.62 + 22.84) / 4 = 19.2475

    Standard deviation, σ = √summation (x - mean) / n

    Summation (x - mean) = (10.31 - 19.2475) ^2 + (17.22 - 19.2475) ^2 + (26.62 - 19.2475) ^2 + (22.84 - 19.2475) ^2 = 151.249475

    σ = √ (151.249475/4)

    σ = 6.15

    number of sample, n = 4

    The z score for 98% confidence interval is 2.33

    We will apply the formula

    Confidence interval

    = mean ± z * standard deviation/√n

    It becomes

    19.2475 ± 2.33 * 6.15/√4

    = 19.2475 ± 2.33 * 3.075

    = 19.2475 ± 7.16

    The lower end of the confidence interval is 19.2475 - 7.16 = 12.09

    The upper end of the confidence interval is 19.2475 + 7.16 = 26.41
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