Question

Plastic Microfiber Pollution on Shorelines

In a study, we see that plastic microparticles are contaminating the world’s shorelines and that much of the pollution appears to come from fibers from washing polyester clothes. The same study also took samples from ocean beaches. Five samples were taken from each of 18 different shorelines worldwide, for a total of 90 samples of size 250 mL. The mean number of plastic microparticles found per 250 mL of sediment was 18.3 with a standard deviation of 8.2 .

(a) Find a 90% confidence interval for the mean number of polyester microfibers per 250 mL of beach sediment.

Round your answers to two decimal places.

The 90% confidence interval is

(b) What is the margin of error?

Round your answer to two decimal places.

margin of error =

(C) If we want a margin of error of only ±1 with 90% confidence, what sample size is needed?

Round your answer up to the nearest integer.

sample size =

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 18.3

Population standard deviation = = 8.2

Sample size = n = 90

a) At 90% confidence level

= 1-0.90% =1-0.90 =0.10

/2 =0.10/ 2= 0.05

Z/2 = Z_{0.05 = 1.645}

Z/2 = 1.645  
Margin of error = E = Z/2 * ( /n)

=1.645 * (8.2 /  90 )

= 1.422

At 90 % confidence interval estimate of the population mean is,

- E < < + E

18.3 - 1.422 <   < 18.3 + 1.422

16.88 <   < 19.72

( 16.88 ,19.72 )

b) Margin of error = E = Z/2 * ( /n)

=1.645 * (8.2 /  90 )

= 1.422

c)

Margin of error = E = 1

Z_/2 = 1.645

sample size = n = [Z_/2* / E] ²

n = [1.645 *8.2 /1 ]²

n = 182

Sample size = n = 182