Respuesta :
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.
