Respuesta :
Answer:
The 95% confidence interval for the population mean amount spent on groceries by Connecticut families is ($73.20, $280.21).
Step-by-step explanation:
The data for the amount of money spent weekly on groceries is as follows:
S = {210, 23, 350, 112, 27, 175, 275, 50, 95, 450}
n = 10
Compute the sample mean and sample standard deviation:
[tex]\bar x =\frac{1}{n}\cdot\sum X=\frac{ 1767 }{ 10 }= 176.7[/tex]
[tex]s= \sqrt{ \frac{ \sum{\left(x_i - \overline{x}\right)^2 }}{n-1} } = \sqrt{ \frac{ 188448.1 }{ 10 - 1} } \approx 144.702[/tex]
It is assumed that the data come from a normal distribution.
Since the population standard deviation is not known, use a t confidence interval.
The critical value of t for 95% confidence level and degrees of freedom = n - 1 = 10 - 1 = 9 is:
[tex]t_{\alpha/2, (n-1)}=t_{0.05/2, (10-1)}=t_{0.025, 9}=2.262[/tex]
*Use a t-table.
Compute the 95% confidence interval for the population mean amount spent on groceries by Connecticut families as follows:
[tex]CI=\bar x\pm t_{\alpha/2, (n-1)}\cdot\ \frac{s}{\sqrt{n}}[/tex]
[tex]=176.7\pm 2.262\cdot\ \frac{144.702}{\sqrt{10}}\\\\=176.7\pm 103.5064\\\\=(73.1936, 280.2064)\\\\\approx (73.20, 280.21)[/tex]
Thus, the 95% confidence interval for the population mean amount spent on groceries by Connecticut families is ($73.20, $280.21).
