Question

The concentration of benzene was measured in units of milligrams per liter for a simple random sample of five specimens of untreated wastewater produced at a gas field. The sample mean was 7.8 with a sample standard deviation of 0.4. Seven specimens of treated wastewater had an average benzene concentration of 6.1 with a standard deviation of 1.7. It is reasonable to assume that both samples come from populations that are approximately normal with unequal variances. Can you conclude at 5% significance that the mean benzene concentration is less in treated water than in untreated water?

(a). What is the null and alternative hypothesis? (The subscripts "T" and "U" denotes treated and untreated water)

• H0:μT=μU,H1:μT≠μU
• H0:¯xT=¯xU,H1:¯xT≠¯xU
• H0:μT=μU,H1:μT<μU
• H0:μT=μU,H1:μT>μU

(b). Compute the test statistic. (round final answer to two decimal places)

(c). Write the R code you would use to compute the p-value.

(d). After running your R code, you find that the p-value is 0.032. You conclude that there is Select an answer insufficient sufficient  evidence that the mean benzene concentration is less in treated water than in untreated water.

(e). A 95% confidence interval for the difference in the mean benzene concentration would contain 0.

True

False

Answer 1

c)

n_u=5
n_t=7
xbar_u=7.8
xbar_t=6.1
s_u=0.4
s_t=1.7
t=2.55
n=round(((s_u^2/n_u)+(s_t^2/n_t))^2/((s_u^4/(n_u^2*(n_u-1)))+(s_t^4/(n_t^2*(n_t-1)))),0)
p_value=2*pt(-abs(t),df=n)

e) Hence False.