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6 March, 20:41

You are interested in determining the effects of a recent coupon promotion on customer purchases at a local grocery store. You sample of 50 customer receipts one day after the coupon is distributed and find the customers' mean spending to be $97.50 with a standard deviation of $4.50. A 90% confidence interval for the population mean would require the use of which critical value: a. t = 1.6766 b. t = 1.2991 c. z = 1.2800 d. z = 1.6450

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  1. 7 March, 00:20
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    (a) t = 1.6766

    Step-by-step explanation:

    n = 50

    degree of freedom = n - 1 = 50 - 1 = 49

    confidence level = 90%

    The critical value corresponding to 90% confidence level and 49 degrees of freedom is obtained by interpolating between 48 and 50 degrees of freedom (df)

    Let the critical value at 49 df be y

    dof. critical value

    48. 1.677

    49. y

    50. 1.676

    (49 - 48) / (50 - 48) = (y - 1.677) / (1.676 - 1.677)

    0.5 * - 1*10^-3 = y - 1.677

    y - 1.677 = - 5*10^-4

    y = - 5*10^-4 + 1.677 = 1.6765

    The closest option is (a)
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