Question

Connecticut families were asked how much they spent weekly on groceries. Using the following data, construct and interpret a 95% confidence interval for the population mean amount spent on groceries (in dollars) by Connecticut families. Assume the data come from a normal distribution

210 23 350 112 27 175 275 50 95 450

Answer 1

The 95% confidence interval for the population mean amount spent on groceries by Connecticut families is ($73.20, $280.21).

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:

`\bar x =(1)/(n)\cdot\sum X=( 1767 )/( 10 )= 176.7`

`s= \sqrt{ \frac{ \sum{\left(x_i - \overline{x}\right)^2 }}{n-1} } = \sqrt{ ( 188448.1 )/( 10 - 1) } \approx 144.702`

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:

`t_(\alpha/2, (n-1))=t_(0.05/2, (10-1))=t_(0.025, 9)=2.262`

*Use a t-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\ (s)/(√(n))`

`=176.7\pm 2.262\cdot\ (144.702)/(√(10))\\\\=176.7\pm 103.5064\\\\=(73.1936, 280.2064)\\\\\approx (73.20, 280.21)`

Thus, the 95% confidence interval for the population mean amount spent on groceries by Connecticut families is ($73.20, $280.21).