A random sample of 200 customers who visited a local grocery spent an average of $97....

A random sample of 200 customers who visited a local grocery spent an average of $97. You find a reputable source stating that the current population standard deviation of all such grocery bills in similar grocery stores is equal to $19. Using this information, find the 90% confidence interval for the population mean:

$95.28 to $98.72
		$94.79 to $99.21
		$94.37 to $99.63
		$93.87 to $100.13

Solutions

Expert Solution

Solution :

Given that,

Point estimate = sample mean = = $97

Population standard deviation = = $19
Sample size = n =200

At 90% confidence level the z is

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

Margin of error = E = Z/2 * ( / $\sqrt$ n)

=1.645 * ( 19 / $\sqrt$ 200 )

= 2.2101
At 90% confidence interval estimate of the population mean
is,

- E < < + E

97 - 2.2101 < <97 + 2.2101

94.79 < < 99.21
$94.79 to $99.21

orchestra answered 1 year ago

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