Question

Plastic microparticles are contaminating the world's shorelines and much of the pollution appears to come from fibers from washing polyester clothes. A study took random samples from ocean beaches. A total of 40 samples of size 250 mL were taken, and the mean number of plastic microparticles of sediment was 18.3 with a standard deviation of 8.2.

1. What is the variable of interest for this situation? Be specific.
2. Check the CLT basic requirements for a confidence interval.
3. Calculate the standard error for a confidence interval and find t* from STATKEY
4. Write the notation for this situation in terms of shape, center, and spread.
5. Calculate the margin of error.
6. Find and interpret a 95% confidence interval for the mean number of polyester microfibers per 250 mL of beach sediment.         

Answer 1

(a) The variable of interest for this situation is the mean number of plastic microparticles of sediment.

(b)

• Random: The data come from a well-designed random sample or randomized experiment.
• Normal: The sampling distribution of the statistic is approximately Normal.
• Independent: Individual observations are independent.

(c) The standard error = s/√n = 8.2/√40 = 1.297

The critical t-value is 2.02.

(d) X ~ N(18.3, 8.2)

(e) The margin of error = t*(s/√n) = 2.02*8.2/√40 = 2.622

(f) The 95% confidence interval for the mean number of polyester microfibers per 250 mL of beach sediment is:

= x t*(s/√n)

= 18.3 2.02*8.2/√40

= (15.678, 20.922)

The 95% confidence interval for the mean number of polyester microfibers per 250 mL of beach sediment is between 15.678 and 20.922.

We are 95% confident that the true mean number of polyester microfibers per 250 mL of beach sediment is between 15.678 and 20.922.

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