Question

7. A researcher wants to know whether the acidity of rain (pH) near Houston, Texas, is significantly different from that near SC Chicago, Illinois. He randomly selects 12 rain dates in Texas and 14 rain dates in Illinois and obtains the following data: Texas 4.69 5.10 5.22 4.46 4.93 4.65 5.22 4.76 4.25 5.14 4.11 4.71 Illinois 4.40 4.22 4.64 4.54 4.35 4.69 4.69 4.75 4.40 4.63 4.45 4.49 4.36 4.52 Source: National Atmospheric Deposition Program (a) Is the sampling method dependent or independent? Why? (b) Because the sample sizes are small, what must be true regarding the populations from which the samples were drawn? (c) Draw side-by-side boxplots of the data. What does the visual evidence imply about the pH of rain in the two states? () Does the evidence suggest that there is a difference in the pH of rain in Chicago and Houston? Use the x = 0.05 level of significance.

Answer 1

Solution

SOLUTION (a) The sampling method is independent, because the samples were randomly selected from independent populations. (b) To use the two-sample t-test, we require that the sampling distributions of the sample mean are approximately normal. Since the samples are small, the central limit theorem cannot be used. The sampling distribution of the sample mean is then approximately normal if the population distribution is approximately normal. Thus we need to assume that the population distributions of both populations) are approximately normal.

Texas Illinois 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 07 5.1 5.2 5.3 5.4 n Mean S Q1 Q3 Max 12 o 0.3539 0.15 4.77 4.5093 4.555 Min 4.11 4.22 0.3696 0.1557 5.12 Median 4.735 4.505 5.22 4.75 14 4.4 4.64

(d) Given claim: difference H1 H2 The claim is either the null hypothesis or the alternative hypothesis. The null hypothesis needs to contain an equality. If the null hypothesis is the claim, then the alternative hypothesis states the opposite of the null hypothesis. HO: M1 = 42 H: Mi + H2

CLASSICAL APPROACH Determine the test statistic: ta 71 - T2 - (11-12) si 2.276 4.77 – 4.5093-0 0.36962 0.15572 12 14 + ni n2 2 ܒ݁ܪܺܝܐ = + 11 n2 Determine the degrees of freedom (rounded down to the nearest integer): si s 0.36962 0.15572 12 14 14 (si/ni)? (sz/n2) (0.3696-/12) (0.15574/14)? ni 12-1 14-1 Determine the critical value from the Student's T distribution table in the appendix in the row with df = 14 and in the column with a/2 = 0.025: 1 n2-1 t = 2.145 The rejection region then contains all values smaller than -2.145 and all values larger than 2.145. If the value of the test statistic is within the rejection region, then the null hypothesis is rejected: 2.276 > 2.145 = Reject Ho There is sufficient evidence to support the claim that there is a difference in the pH of rain in Chicago and Houston.

P-VALUE APPROACH Determine the test statistic: t- 71 - T2 - (11-12) si 2.276 4.77 – 4.5093-0 0.36962 0.15572 12 14 + ni n2 2 + 11 n2 Determine the degrees of freedom (rounded down to the nearest integer): si s 0.36962 0.15572 12 14 14 (s/n)(s/n2) (0.3696-/12) (0.15574/14) 1 12-1 14-1 The P-value is the probability of obtaining the value of the test statistic, or a value more extreme, assuming that the null hypothesis is true. The P-value is the number or interval) in the column title of the Students T distribution in the appendix containing the t-value in the row df = 14 (Note: We double the boundaries, because the test is two-tailed) ni n2-1 0.02 = 2 x 0.01 <P < 2 x 0.02 = 0.04 If the P-value is less than the significance level, reject the null hypothesis. P<0.05 → Reject Ho There is sufficient evidence to support the claim that there is a difference in the pH of rain in Chicago and Houston.