Convert the adjusted sample proportions into Zscores within

Convert the adjusted sample proportions into Z-scores within probability notation. Use 4th decimal accuracy.

Round the Z-scores to the nearest 2nd decimal (for Z-table use) within probability notation. Arrange the notation so that we are subtracting the lower bounded probability from the upper bounded probability.

Replace the upper and lower bounded probability notations with their respective probabilities found in the Z-table. Use 6 decimal accuracy.

Show your final answer after subtraction, and summarize with an English sentence with an answer rounded to the nearest percentile.

About ____% of the time left-handers will make up between 8 to 12% of a random sample of 29 people.

Solution

p0 = 0.09
n = 29

standard error, SE = sqrt(p0(1-p0)/n)
SE = sqrt(0.09(1-0.09) / 29) = 0.05314

P(0.08 < phat< 0.12)
= P( ((0.08-0.09) / 0.053) <Z< ((0.12-0.09) / 0.053) )
= P( -0.19 <Z< 0.57 )
= P(Z<0.57) - P(Z<-0.19)
= P(Z<0.57) - (1 - P(Z<0.19))
= 0.7157 - ( 1 - 0.5753)
= 0.7157 - 0.4247
= 0.2910

About 29% of the time left-handers will make up between 8 to 12% of a random sample of 29 people.