In an examination of purchasing patterns of shoppers, a sample of 36 shoppers revealed that they spent, on average, $50 per hour of shopping. Based on previous years, the population standard deviation is $4.80 per hour of shopping. Assuming that the amount spent per hour of shopping is normally distributed, calculate a 90% confidence interval.

the 90% confidence interval is ( 48.684  , 51.316  )

Step-by-step explanation:

Given that :

the sample size = 36

Sample Mean = 50

standard deviation = 4.80

The objective is to calculate a 90% confidence interval.

At 90% confidence interval ;

the level of significance = 1 - 0.9 = 0.1

The critical value for

= 1.645

The standard error S.E =

=

= 0.8

The Confidence interval level can be computed as:

For the lower limit :

=50 - 1.316

= 48.684

For the upper limit :

=50 + 1.316

= 51.316

Thus, the 90% confidence interval is ( 48.684  , 51.316  )

idk

step-by-step explanation: