Question
In: Statistics and Probability
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
* ( 
/
=1.645 * ( 19 /
= 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 2 years agoRelated Solutions
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
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