Question

Carbon monoxide concentrations (in µg/m3 ) are measured on 6 different days in city A and on 4 different days in city B. The following measurements have been obtained in cities A and B, respectively: 3.9, 4.7, 7.1, 6.9, 4.3, 6.3 and 7.3, 6.9, 7.6, 9.1. We assume that these two samples are independent.

(a) Assume that both samples are from normally distributed populations. We also assume that the same type of measurement device was used in both cities so that the measurement error (variance) is the same for both samples. Construct the 95% confidence interval for δ = µA−µB where µA and µB are mean concentrations of carbon monoxide in cities A and B, respectively. (b) Do we reject H0 : δ = 0 in favor of H1 : δ 6= 0 at the 5% significance level? Why or why not? Find the p-value of this test.

(c) Now assume that both samples are from normally distributed populations, as before, but two different measurement devices were used and their measurements errors are known. We assume that the variances are 2 and 1, respectively, for the two samples. Construct the 95% confidence interval for δ.

(d) Compare confidence intervals you obtained in (a) and (c). Which one is narrower? Briefly explain why.

(e) Show that (3.9, 7.1) is an approximate 97% confidence interval for the median concentration of carbon monoxide in city A.

Answer 1

(a)

Independent samples with equal variances:

City A:

Sample size, n1 =6

Sample mean, =5.533

Sample variance, s1² =1.9587

City B:

Sample size, n2 =4

Sample mean, =7.725

Sample variance, s2² =0.9225

Pooled std.deviation, S_p = = =1.2530

Standard Error, SE =S_p* =1.2530* =0.8088

The critical value of t at df =n1+n2-2 =6+4-2 =8, for a two-tailed test at 0.05 significance level is: t_crit =2.3060

Margin of Error, MoE =t_crit*SE =2.3060*0.8088 =1.8651

The 95% confidence interval for δ = µA−µB =( =(5.533 - 7.725) 1.8651 =(-4.0571, -0.3269)

(b)

Since the 95% confidence interval (-4.0571, -0.3269) does not contain the value 0, we reject the null hypothesis(H0) at the 5% significance level in favor of the alternative hypothesis(H1).

Test statistic, t =( =(5.533 - 7.725)/0.8088 = -2.710

The p-value for t = -2.7102 and df =n1+n2-2 =8 is p-value =0.027

(c)

Independent samples with unequal variances:

Standard Error, SE = = =0.7638

Calculation of degrees of freedom(df):

At df =8 and at 5% significance level, for a two-tailed case, t_crit =2.3060

Margin of Error, MoE =t_crit*SE =2.3060*0.7638 =1.7613

The 95% confidence interval for δ =(​​​​​​)MoE =(5.533 - 7.725) 1.7613 =(-3.9533, -0.4307)

(d)

Comparing the confidence intervals in (a) and (c):

The confidence interval in (c) is narrower than that of (a) because (c) has lower margin of error (1.7613) than that of (a) (1.8651).