Respuesta :
Answer:
The 95% confidence interval for the mean savings is ($60.54, $81.46).
Step-by-step explanation:
As there is no information about the population standard deviation of savings and the sample is not large, i.e. n = 20 < 30, we will use a t-confidence interval.
The (1 - α)% confidence interval for population mean (μ) is:
[tex]CI=\bar x\pm t_{\alpha/2, (n-1)}\times \frac{s}{\sqrt{n}}[/tex]
For the data provided compute the sample mean and standard deviation as follows:
[tex]\bar x=\frac{1}{n}\sum\limits^{n}_{i=1}{ X_{i}}=\frac{1}{20}\times [92+34+40+...+53+82]=71[/tex]
[tex]s= \sqrt{ \frac{ \sum{\left(x_i - \overline{X}\right)^2 }}{n-1} } = \sqrt{ \frac{ 9492 }{ 20 - 1} } \approx 22.35[/tex]
The critical value of t for α = 0.05 and (n - 1) = 19 degrees of freedom is:
[tex]t_{\alpha/2, (n-1)}=t_{0.025, 19}=2.093[/tex]
*Use a t-table for the value.
Compute the 95% confidence interval for the mean savings as follows:
[tex]CI=\bar x\pm t_{\alpha/2, (n-1)}\times \frac{s}{\sqrt{n}}\\=71\pm 2.093\times\frac{22.35}{\sqrt{20}}\\=71\pm 10.46\\=(60.54, 81.46)[/tex]
Thus, the 95% confidence interval for the mean savings is ($60.54, $81.46).
