A survey of 50 retail stores revealed that the average price of a microwave was $375 with a sample standard deviation of $20. Assuming the population is normally distributed, what is the 99% confidence interval to estimate the true cost of the microwave

The 99% C.I is ( 367.703, 382.297 )

Step-by-step explanation:

Given that:

sample mean = 375

Population standard deviation = 20

Sample size n = 50

The level of significance =  1 - C.I = 1 - 0.99 = 0.01

From z tables:

The critical value = = = 2.58

The margin of error E =

The margin of error E =  

The margin of error E = 7.297

The limits of 99% C.I are expressed as:

Lower limit = = 375 - 7.297

Lower limit = 367.703

Upper limit = =  375 + 7.297

Upper limit = 382.297

The 99% C.I is =

= ( 367.703, 382.297 )

none.

step-by-step explanation:

m

1

=

5

m1=5

b

=

−

1

b=-1

find the slope and y-intercept of the second equation.

tap for more

m

2

=

0

m2=0

b

=

−

6

b=-6

compare the slopes

m

m of the two equations.

m

1

=

5

,

m

2

=

0

m1=5,m2=0

compare the two slopes in the decimal form. if the slopes are equal, then the lines are parallel. if the slopes are not equal, then the lines are not parallel.

m

1

=

5

,

m

2

=

0

m1=5,m2=0

(x) - (y) = (1 + 4i)