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.

Answers

Answer 1

Answer:

a) 70,663.57< [tex]\overline x[/tex] < 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, [tex]\overline x[/tex] = $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

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

[tex]t^*[/tex] = The critical t, at 95% confidence level = 1.67

Therefore, we have;

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

Therefore, we have;

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

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< [tex]\overline x[/tex] < 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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Step-by-step explanation:

Peter is the owner of a hardware supply shop. He has hired you to check his machines calibration prior to starting production on a large order. To check this, you set the machine to create 2.0 inch nails and manufacture a random sample of 200 nails. That sample of nails has an average length of 2.024 inches with a standard deviation of 0.205 inches.


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Answers

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Step-by-step explanation:

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How to calculate the test statistics.

For the null hypothesis, we would test that:

H0 : p = 2

For the alternate hypothesis, we would test that:

Ha > p = 2

In Mathematics, the test statistics of a given sample is calculated by using this formula:

[tex]Z=\frac{x\;-\;u}{\frac{\sigma}{\sqrt{n} } }[/tex]

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[tex]Z=\frac{2.024\;-\;2}{\frac{0.205}{\sqrt{200} } }\\\\Z=\frac{0.024}{\frac{0.205}{14.14} }\\\\Z=\frac{0.024}{0.0145 }[/tex]

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Answers

Answer:

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Step-by-step explanation:

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