Question

• Calculate means and standard deviations of these percentages for each country based on sample size = 20.
• Test at 5% level individually if the mean percentage for each of the countries

Book 5 - Saved View Help Tell me what you want to do Open in Desktop App Review = = Percentage E P E À Autosum. 47 E Insert $ Delete Format % 9 Conditional Formatting 0 Format as Table -48 Clear Sort & Filter Alignment Number Tables Cells har Editing G H I J M N O P Q Peru deaths per Total cases % -00% ..00% .00% 17 1.50% 234 671 0.30% .20% .40% 0.30% 0.50% 0.70% 0.90% 0.90% 0.90% 1.00% 0.80% 0.80% 0.90% 0.90% 0.90% 0.90% Dates Deaths Cases Total Deaths per Total cases % 16-Mar 86 0.00% 117 0.00% 145 0.00% 0.40% 318 1.50% 395 1.30% 635 1.70% 2.40% 950 2.50% 2.80% 1-Apr 1323 2.80% 1414 3.90% 1595 13.80% 17464.10% 121 4342 2.70% 5256 2.60% 169 58972.90% 6848 2.60% 10303 2.20% 12491 2.20% 20 ooo 181 20 PrtScn Home End Fio PgUP FI % &

is different from 10%.
• Obtain a 95% confidence interval for the difference of means for the two countries.
• Test an appropriate pair of hypotheses for the two means at 5% level of significance.

sorry about the placement of pictures!! they're not meant to be in between the bullet points

Answer 1

Suppose random variables X and Y denote death percentages in different days in Peru and Russia respectively. (a) In case of Peru, Sample size ny = 20 Mean percentage is given by T =- į r;= 2.12 Standard deviation is given by 8=1-13 (–3) = 1 I (:3 – .12)2 = 1.250741 In case of Russia, Sample size ny = 20 Mean percentage is given by

120 jay ny 21 = yi = 0.59 20 Standard deviation is given by (yi - )2 = (yi - 0.59)2 = 0.3522858 (b) In case of Peru, We have to test for null hypothesis Ho: My = 10 against the alternative hypothesis H1 :: 7 10 Our test statistic is given by - FLO s/vn, Here, ny = 20

ī = 2.12 8, = 1.259741 2.12 - 10 . tcalculated = 1.259741/20 = -27.97435 Degrees of freedom U, = n; -1 = 19 p-value = P(t19 < -27.97435) + P (t19 > 27.97435) = 3.321045 * 10-17 [Using R-code '1-pt(27.97435,19)+pt(-27.97435,19)'] Level of significance a = 0.05 We reject our null hypothesis if p-value <a Here, we observe that p-value = 3.321045 * 10-17 < 0.05 = a So, we reject our null hypothesis. Hence, based on the given data we can conclude that the percentage for Peru is significantly different from 10%. In case of Russia, We have to test for null hypothesis Ho: My = 10

against the alternative hypothesis H . Hy + 10 Our test statistic is given by + * Y - LLO sul ny Here, ny = 20 y = 0.59 Sy = 0.3522858 0.59 - 10 . tcalculated = 1 == -119.4564 0.3522858/20 Degrees of freedom Wy =ny -1 = 19 p-value = P(t19 < -119.4564) + P (019 > 119.4564) = 4.283249 * 10-29 [Using R-code '1-pt(119.4564,19)+pt(-119.4564,19)'] Level of significance a = 0.05 We reject our null hypothesis if p-value <a Here, we observe that P-value = 4.283249 * 10-29 < 0.05 = a

So, we reject our null hypothesis. Hence, based on the given data we can conclude that the percentage for Russia is significantly different from 10% (c) Degrees of freedom v = n, tn, - 2 = 38 We know, P(-2.024394 <t38 < 2.024394) = 0.95 = P(-2.02894 <17, < 2.02494) = 0,95 = P(-2.024394< 0.9249467 + V1/20+1/20<2 (2.12 -0.59) - d o < 2.024394) = 0.95 = P(-2.024391 < 0.563,49% < 2.024394) = 0,95 = P(-2.024394 * 0.2924938< 1.53 - d < 2.024394 * 0.2924938) = 0.95 → P(1.53 -0.5921227< d <1.53+0.5921227) = 0.95 ::P(0.9378773 < d < 2.122123) = 0.95 Hence, 95% confidence interval for the difference of means of two countries is given by (0.9378773, 2.122123)

(d) Here, two different groups are used to collect data in two different countries. Further we do not know population standard deviation (or variance). So, we have to perform two sample t-test. We have to test for null hypothesis How = My against the alternative hypothesis H1:, My Our test statistic is given by S* V1/n, +1/ny Here, n; = 20 Ny = 20 I = 2.12 y = 0.59 /nz - 1)*sz + (ny - 1)* **= 0.9249467 _ ny + ny - 2 2.12 -0.59 .. calculated 09249467 * * V1/20+ 1/20 = 5.23088 1/20+ 1/20

Degrees of freedom v = n, tn, -2 = 38 p-value = P(+38 > 5.23088) + P (t38 < 5.23088) = 1.831073 * 10-17 [Using R-code '1-pt(5.23088,38)-pt(-5.23088,38)] Level of significance a = 0.05 We reject our null hypothesis if P-value <a Here, we observe that p-value = 1.831073 * 10-17 < 0.05 = a So, we reject our null hypothesis. Hence, based on the given data we can conclude that mean percentages of Peru and Russia are significantly different