Comparing two groups #2 - Questions and Answers

Question 1 The population 𝑃 P (in thousands) of Tallahassee, Florida, from 2000 through 2014 can be modeled by 𝑃 = 150.9 × 𝑒 𝑘 𝑡 P=150.9×e kt , where 𝑡 t represents the year, with 𝑡 = 0 t=0 corresponding to 2000. In 2005, the population of Tallahassee was about 163,075. (a) Find the value of 𝑘 k. Is the population increasing or decreasing? Explain. (b) Use the model to predict the populations of Tallahassee in 2020 and 2025. Are the results reasonable? Explain. (c) According to the model, during what year will the population reach 200,000? Answer Step 1 Given: The population model is given as, 𝑃 = 150.9 × 1000 × 𝑒 𝑘 𝑡 P=150.9×1000×e kt (a) To find the value of 𝑘 k, multiply ln on both sides. 163075 = 150.9 × 1000 × 𝑒 5 𝑘 163075=150.9×1000×e 5k ln ⁡ 163075 = ln ⁡ ( 150900 × 𝑒 5 𝑘 ) ln163075=ln(150900×e 5k ) ln ⁡ 163075 = ln ⁡ ( 150900 ) + 5 𝑘 ln163075=ln(150900)+5k 𝑘 = ln ⁡ ( 163075 ) − ln ⁡ ( 150900 ) 5 k= 5 ln(163075)−ln(150900) ​ 𝑘 = ln ⁡ ( 1.0776 ) 5 k= 5 ln(1.0776) ​ 𝑘 = 0.01552 k=0.01552 Since 𝑘 > 0 k>0, the value of 𝑘 k is positive. Hence the population is increasing. Step 2 (b) The population in 2020 is, 𝑡 = 2020 − 2000 t=2020−2000 𝑡 = 20 t=20 𝑃 = 150.9 × 1000 × 𝑒 𝑘 𝑡 P=150.9×1000×e kt 𝑃 = 150900 × 𝑒 0.01552 × 20 P=150900×e 0.01552×20 𝑃 = 150900 × ( 1.3634 ) P=150900×(1.3634) 𝑃 = 205740 P=205740 The population in 2025 is, 𝑡 = 2025 − 2000 t=2025−2000 𝑡 = 25 t=25 𝑃 = 150.9 × 1000 × 𝑒 𝑘 𝑡 P=150.9×1000×e kt 𝑃 = 150900 × 𝑒 0.01552 × 25 P=150900×e 0.01552×25 𝑃 = 150900 × ( 1.4622 ) P=150900×(1.4622) 𝑃 = 220658 P=220658 By comparing the population in 2005, the population in 2020 and 2025 is very low amount of growth. Hence the obtained results are not reasonable. Step 3 (c) To find the year at which the population is 200,000: 𝑃 = 150.9 × 1000 × 𝑒 𝑘 𝑡 P=150.9×1000×e kt 200000 = 150900 × 𝑒 0.01552 𝑡 200000=150900×e 0.01552t 200000 150900 = 𝑒 0.01552 𝑡 150900 200000 ​ =e 0.01552t Now taking ln on both sides, ln ⁡ 200000 150900 = ln ⁡ ( 𝑒 0.01552 𝑡 ) ln 150900 200000 ​ =ln(e 0.01552t ) 0.2817 = 0.01552 𝑡 0.2817=0.01552t 𝑡 = 0.2817 0.01552 t= 0.01552 0.2817 ​ 𝑡 = 18.151 t=18.151 Hence in a time period of 𝑡 ≈ 18 t≈18 years the population will be 200,000. Question 2 Describe the basic characteristics that define an independent-measures, or a between-subjects, research study. Answer An independent measures design is the one in which different participants are used in each condition of the independent variable. Different participants are used for the two groups i.e. every participant is only subjected to a single treatment. Since independent measure design uses two distinct populations (and population means), hence, each of these have their own sample mean and variances. Question 3 To explain: When the relative frequencies are useful in comparing two data sets. Answer Relative frequency distribution: Relative frequency distribution is the frequency distribution in which the proportion of an event happening is considered, rather than the number of happenings of that event. If the relative frequencies for all events in the data are listed, then it is relative frequency distribution. The general formula for the relative frequency: Relative frequency = Frequency of the data Total frequency Relative frequency= Total frequency Frequency of the data ​ Relative frequencies are useful in comparing two different data sets to the whole population. Frequencies are used to compare different data sets and different classes, but not with the entire data set and not with the entire population. If the size of the two samples or populations is different, then the relative frequencies are used in comparing two data sets. Question 4 Provide three examples of studies when we can use (a) the pooled, (b) non-pooled, and (c) paired inference to compare means of two populations. Answer In a problem of comparing means of two populations we may come across the following conditions depending on the property of data. (a) Pooled variance: when the data selected for comparing two population variance. Then we used pooled variance to draw inferences and compare the means. It 𝑠 1 2 s 1 2 ​ and 𝑠 2 2 s 2 2 ​ are the sample variances of two population with equal population variances, then "pooled variance" is given by: 𝑠 𝑝 2 = ( 𝑛 1 − 1 ) 𝑠 1 2 + ( 𝑛 2 − 1 ) 𝑠 2 2 𝑛 1 + 𝑛 2 − 2 s p 2 ​ = n 1 ​ +n 2 ​ −2 (n 1 ​ −1)s 1 2 ​ +(n 2 ​ −1)s 2 2 ​ ​ In this case, the test statistic for comparing means is given by: 𝐴 = 𝑥 1 ‾ − 𝑥 2 ‾ 𝑠 𝑝 1 𝑛 1 + 1 𝑛 2 A= s p ​ n 1 ​ 1 ​ + n 2 ​ 1 ​ ​ x 1 ​ ​ − x 2 ​ ​ ​ (b) Non-pooled variance: On the contrary, the property discussed in part (a) about the data, when the populations variances of two populations are assumed to be unequal, then non-pooled variance is used to compare the means of two population. It is given as 𝑆 𝑛 𝑜 𝑛 − 𝑝 𝑜 𝑜 𝑙 𝑒 𝑑 2 = 𝑠 1 2 𝑛 1 + 𝑠 2 2 𝑛 2 S non−pooled 2 ​ = n 1 ​ s 1 2 ​ ​ + n 2 ​ s 2 2 ​ ​ and the test statistic is 𝐴 = 𝑥 1 ‾ − 𝑥 2 ‾ 𝑠 1 2 𝑛 1 + 𝑠 2 2 𝑛 2 A= n 1 ​ s 1 2 ​ ​ + n 2 ​ s 2 2 ​ ​ ​ x 1 ​ ​ − x 2 ​ ​ ​ (c) Paired inference (Paired test): Let us consider the case when: (i) Sample sizes are equal and (ii) The samples are not independent but the sample observations are paired together. i.e., the pair of observations ( 𝑥 𝑖 , 𝑦 𝑖 )   ( 𝑖 = 1 , 2 , . . . 𝑛 ) (x i ​ ,y i ​ ) (i=1,2,...n) correspond to the same n-th unit. For example: Suppose we want to test the efficacy of a particular drug, say, for inducing sleep. Let 𝑥 𝑖 x i ​ and 𝑦 𝑖   ( 𝑖 = 1 , 𝑛 ) y i ​  (i=1,n) be the readings, in hours of sleep, on the i-th individual before and after the drug is given respectively.