Finding new customers An online retailer wants to estimate p = the proportion of all orders in a given week that are from new customers.

Question

Finding new customers An online retailer wants to estimate p = the proportion of all orders in a given week that are from new customers. He takes an SRS of 40 orders and finds that 12 of them are from new customers. Calculate a 99% confidence interval for the proportion of the week’s orders that are from new customers.

{A} .11 to .49

{B} .18 to .33

{C} .2 to .4

{D} .25 to .45

Answer 1

Favorable Cases X =

12

Sample Size N =

40

The sample proportion is computed as follows, based on the sample size N=40 and the number of favorable cases X=12:

p^{^}​=N/X​=4012​=0.3

The critical value for α=0.01 is zc​=z(1−α/2)​=2.576. The corresponding confidence interval is computed as shown below:

CI(Proportion)​===​(p​−zc ((sqrt​(p^​(1−p^​)​​/n) , p+zc(sqrt(p^​(1−p^​)/n​​))

=(0.3−(2.576× sqrt((0.3(1−0.3)​​/40)) ,0.3+2.576× sqrt((0.3(1−0.3)​​/40))

=(0.113,0.487)​

Therefore, based on the data provided, the 99% confidence interval for the population proportion is 0.113<p<0.487, which indicates that we are 99% confident that the true population proportion p is contained by the interval (0.113,0.487).

Round off to two decimal it will become 0.11 and 0.49

=

so Answer for this question option. A