Question

Let x be a random variable that represents micrograms of lead per liter of water (µg/L). An industrial plant discharges water into a creek. The Environmental Protection Agency (EPA) has studied the discharged water and found x to have a normal distribution, with

σ = 0.7 µg/L.

† Note: For degrees of freedom d.f. not in the Student's t table, use the closest d.f. that is smaller. In some situations, this choice of d.f. may increase the P-value a small amount and thereby produce a slightly more "conservative" answer.

(a) The industrial plant says that the population mean value of x is

μ = 2.0 µg/L.

However, a random sample of

n = 10

water samples showed that

x = 2.53 µg/L.

Does this indicate that the lead concentration population mean is higher than the industrial plant claims? Use

a = 1%.

(i) What is the level of significance?

What is the value of the sample test statistic? (Round your answer to two decimal places.)

(b) Find a 95% confidence interval for μ using the sample data and the EPA value for σ. (Round your answers to two decimal places.)

lower limit	µg/L
upper limit	µg/L

(c) How large a sample should be taken to be 95% confident that the sample mean

x

is within a margin of error

E = 0.4 µg/L

of the population mean? (Round your answer up to the nearest whole number.)
water samples

Answer 1

The statistical software output for this problem is:

One sample Z summary hypothesis test:
μ : Mean of population
H₀ : μ = 2
H_A : μ > 2
Standard deviation = 0.7

Hypothesis test results:

Mean	n	Sample Mean	Std. Err.	Z-Stat	P-value
μ	10	2.53	0.22135944	2.3942959	0.0083

95% confidence interval results:

Mean	n	Sample Mean	Std. Err.	L. Limit	U. Limit
μ	10	2.53	0.22135944	2.0961435	2.9638565

Hence,

a) i) Level of significance = 0.01

Sample test statistic = 2.39

b) 95% confidence interval:

Lower limit = 2.10 mg/L

Upper limit = 2.96 mg/L

c) Sample size

n = 12