Question

 

Lead in amounts over the primary drinking water standard of 0.0150 milligrams per liter may cause nervous system disorders and brain or kidney damage. Since lead accumulates in body tissue, it is especially hazardous to the fetus or to children under three years old. Over time, the standard deviation of the lead concentration in samples of the drinking water of a neighborhood near an old battery plant has been s=0.0005 mg/l. This week, 16 samples of drinking water were collected from homes in the neighborhood with a mean of =0.0153.

a) Do these statistics provide sufficient evidence to require residents to stop drinking the water and to incur the expense of having water trucked into the neighborhood? Do a hypothesis test and show all 6 steps. Use.

b) Briefly discuss the consequences of type I and type II errors in the context of this problem. From your point of view, which of the two errors is the most serious for this situation?

c) Construct a 95% confidence interval for the mean amount of lead in the water.  confidence interval

d) What is the power of the test for an alternative mean of 0.0154?

Answer 1

a)

State the hypotheses.

Null hypothesis H0: = 0.0150 milligrams per liter

Alternative hypothesis Ha: > 0.0150 milligrams per liter

Formulate an analysis plan. For this analysis, the significance level is 0.05. Since we know the population standard deviation, we will use one-sample z-test.

Critical values -

For significance level of 0.05, the critical value of z is 1.96

Analyze sample data.

Standard error of mean = s / = 0.0005 / = 0.000125

Test statistic, z = (Observed mean - hypothesized mean) / Standard error

= (0.0153 - 0.0150) / 0.000125

= 2.4

P-value = P(z > 2.4) = 0.008

Decision -

Since the test statistic is greater than the critical value (or p-value is less than the significance level), we reject the null hypothesis H0.

Conclusion -

We reject null hypothesis H0 and conclude that there is significant evidence that mean lead content is greater than 0.0150 milligrams per liter.

b)

Type | error (rejection of a true null hypothesis) - We conclude that mean lead content is greater than 0.0150 milligrams per liter but in reality that mean lead content is 0.0150 milligrams per liter.

Type |I error (failure to reject a false null hypothesis) - There is no significant evidence that mean lead content is greater than 0.0150 milligrams per liter but in reality that mean lead content is greater than 0.0150 milligrams per liter.

c)

95% confidence interval for the mean amount of lead in the water is,

(0.0153 - 1.96 * 0.000125,  0.0153 + 1.96 * 0.000125)

= (0.015055, 0.015545)

d)

Critical value of mean to reject H0 = 0.0150 + 1.96 * 0.000125 = 0.015245

Power of the test for an alternative mean of 0.0154 = P(Reject H0 | = 0.0154)

= P(mean > 0.015245 |   = 0.0154)

= P(Z > (0.015245 - 0.0154) / 0.000125)

= P(Z > -1.24)

= 0.8925