Business bankruptcies in Canada are monitored by the Office of the Superintendent of Bankruptcy Canada (OSB).8 Included in each report are t

Question

Business bankruptcies in Canada are monitored by the Office of the Superintendent of Bankruptcy Canada (OSB).8 Included in each report are the assets and liabilities the company declared at the time of the bankruptcy filing. A study is based on a random sample of 75 reports from the current year. The average debt (liabilities minus assets) is $92,172 with a standard deviation of $111,538.
a) Construct a 95% one-sample t confidence interval for the average debt of these companies at the time of filing.
b) Because the sample standard deviation is larger than the sample mean, this debt distribution is skewed. Provide a defense for using the tconfidence interval in this case.

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bonexptip 1 day 2021-07-22T06:42:18+00:00 1 Answers 0 views 0

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    2021-07-22T06:44:06+00:00

    Answer:

    a) 70,663.57< \overline x < 113,680.43

    b) The sampling distribution of the means is expected to be approximately normal given that n > 30

    Step-by-step explanation:

    The given parameter of the study are;

    The number of reports in the study, n = 75 reports

    The average debt, \overline x = $92,172

    The standard deviation, Sₓ = $111,538

    a) A one-sample t confidence interval is given as follows;

    The Degrees of Freedom, df = n – 1 = 75 – 1 = 74

    C.I, = \overline x \pm t^* \cdot \dfrac{S_x}{\sqrt{n} }

    t^* = The critical t, at 95% confidence level = 1.67

    Therefore, we have;

    C.I, = 92,172 \pm 1.67 \cdot \dfrac{111,538}{\sqrt{75} }

    Therefore, we have;

    C.I, = 92,172 \pm 1.67 \cdot \dfrac{111,538}{\sqrt{75} }

    C.I. = (92,172 ± 21,508.43)

    Therefore, the 95% one-sample t confidence interval for the average debt of the companies at the time of filing, C.I. = 70,663.57< \overline x < 113,680.43

    b) The defense for using the t confidence interval in the question is supported that the fact that the sample size is larger than 30 (n = 75) the sampling distribution of means will be approximately normal and the population mean will be between the confidence interval for the mean.

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