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 = 2.55 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? State the null and alternate hypotheses. H0: μ = 2.0; H1: μ ≠ 2.0 H0: p = 2.0; H1: p < 2.0 H0: p = 2.0; H1: p > 2.0 H0: μ = 2.0; H1: μ > 2.0 H0: μ = 2.0; H1: μ < 2.0 H0: p = 2.0; H1: p ≠ 2.0 (ii) What sampling distribution will you use? What assumptions are you making? The standard normal, since we assume that x has a normal distribution with known σ. The Student's t, since we assume that x has a normal distribution with unknown σ. The Student's t, since we assume that x has a normal distribution with known σ. The standard normal, since we assume that x has a normal distribution with unknown σ. What is the value of the sample test statistic? (Round your answer to two decimal places.) (iii) Find (or estimate) the P-value. P-value > 0.500 0.250 < P-value < 0.500 0.100 < P-value < 0.250 0.050 < P-value < 0.100 0.010 < P-value < 0.050 P-value < 0.010 Sketch the sampling distribution and show the area corresponding to the P-value. WebAssign Plot WebAssign Plot WebAssign Plot WebAssign Plot (iv) Based on your answers in parts (i) to (iii), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level α? At the α = 0.01 level, we reject the null hypothesis and conclude the data are statistically significant. At the α = 0.01 level, we reject the null hypothesis and conclude the data are not statistically significant. At the α = 0.01 level, we fail to reject the null hypothesis and conclude the data are statistically significant. At the α = 0.01 level, we fail to reject the null hypothesis and conclude the data are not statistically significant. (v) Interpret your conclusion in the context of the application. There is sufficient evidence at the 0.01 level to conclude that the population mean discharge level of lead is higher. There is insufficient evidence at the 0.01 level to conclude that the population mean discharge level of lead is higher. (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

a)

i) level of significance =0.01

H0: μ = 2.0; H1: μ > 2.0

ii)

The standard normal, since we assume that x has a normal distribution with known σ

population mean μ=		2
sample mean 'x̄=		2.550
sample size    n=		10
std deviation σ=		0.70
std error ='σx=σ/√n=0.7/√10=		0.221
z statistic= ='(x̄-μ)/σx=(2.55-2)/0.221=		2.48

iii)

P-value < 0.010

iv()

At the α = 0.01 level, we reject the null hypothesis and conclude the data are statistically significant.

There is sufficient evidence at the 0.01 level to conclude that the population mean discharge level of lead is higher.

b)

for 95 % CI value of z=		1.960	from excel:normsinv((1+0.95)/2)
margin of error E=z*std error =		0.4339
lower bound=sample mean-E=		2.12
Upper bound=sample mean+E=		2.98

c)

for95% CI crtiical Z          =	1.96	from excel:normsinv(0.975)
standard deviation σ=	0.700
margin of error E =	0.4
required n=(zσ/E)2 =	12	Rounding up