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}
<em>n</em> = 10
Compute the sample mean and sample standard deviation:
![\bar x =\frac{1}{n}\cdot\sum X=\frac{ 1767 }{ 10 }= 176.7](https://tex.z-dn.net/?f=%5Cbar%20x%20%3D%5Cfrac%7B1%7D%7Bn%7D%5Ccdot%5Csum%20X%3D%5Cfrac%7B%201767%20%7D%7B%2010%20%7D%3D%20176.7)
![s= \sqrt{ \frac{ \sum{\left(x_i - \overline{x}\right)^2 }}{n-1} } = \sqrt{ \frac{ 188448.1 }{ 10 - 1} } \approx 144.702](https://tex.z-dn.net/?f=s%3D%20%5Csqrt%7B%20%5Cfrac%7B%20%5Csum%7B%5Cleft%28x_i%20-%20%5Coverline%7Bx%7D%5Cright%29%5E2%20%7D%7D%7Bn-1%7D%20%7D%09%09%09%09%09%09%20%3D%20%5Csqrt%7B%20%5Cfrac%7B%20188448.1%20%7D%7B%2010%20-%201%7D%20%7D%20%5Capprox%20144.702)
It is assumed that the data come from a normal distribution.
Since the population standard deviation is not known, use a <em>t</em> confidence interval.
The critical value of <em>t</em> for 95% confidence level and degrees of freedom = n - 1 = 10 - 1 = 9 is:
![t_{\alpha/2, (n-1)}=t_{0.05/2, (10-1)}=t_{0.025, 9}=2.262](https://tex.z-dn.net/?f=t_%7B%5Calpha%2F2%2C%20%28n-1%29%7D%3Dt_%7B0.05%2F2%2C%20%2810-1%29%7D%3Dt_%7B0.025%2C%209%7D%3D2.262)
*Use a <em>t</em>-table.
Compute the 95% confidence interval for the population mean amount spent on groceries by Connecticut families as follows:
![CI=\bar x\pm t_{\alpha/2, (n-1)}\cdot\ \frac{s}{\sqrt{n}}](https://tex.z-dn.net/?f=CI%3D%5Cbar%20x%5Cpm%20t_%7B%5Calpha%2F2%2C%20%28n-1%29%7D%5Ccdot%5C%20%5Cfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D)
![=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)](https://tex.z-dn.net/?f=%3D176.7%5Cpm%202.262%5Ccdot%5C%20%5Cfrac%7B144.702%7D%7B%5Csqrt%7B10%7D%7D%5C%5C%5C%5C%3D176.7%5Cpm%20103.5064%5C%5C%5C%5C%3D%2873.1936%2C%20280.2064%29%5C%5C%5C%5C%5Capprox%20%2873.20%2C%20280.21%29)
Thus, the 95% confidence interval for the population mean amount spent on groceries by Connecticut families is ($73.20, $280.21).