Question

Q13. The sample proportions p₁ and p₂ are pooled to create a single value for testing

          H₀: π₁=π₂ (say π) vs some suitable H₁. They are pooled because

          a. adding sample sizes makes a still larger size to make Z test valid and efficient

         b. pooling proportions of unequal sample sizes increases efficiency of Z to favor H₀.

          c. pooled proportion satisfies all characteristics of a, b, d

          d. H₀ is assumed true under which pooled proportion will be unbiased for π

Q14. In a two tail test, the probability P(Z_STAT ≤ -2.12)=0.017. The true statement (s) is/are:

a. Conclusions reached in b, c, d are true

b. the null hypothesis should not be rejected at α=0.02

c. the null hypothesis should be rejected at α=0.04

d. an alternative should be rejected at α=0.02

Q15. From studies it is known that the more the time spent running on a treadmill the more the

         calories will be burnt. Let ρ (rho) denotes the population linear correlation coefficient between

         time spent running on treadmill and calories burnt in the gym then for H₀: ρ=0 vs H₁: ρ≠0 has

         to be verified to support the previous studies. To verify this relationship a random sample of 27

         individuals from gym is selected. Given α (alpha) = 0.05, the minimum simple linear correlation

         coefficient (r) that will reject H₀ will be

         a. 0.3819

         b. 0.3809

         c. 0.3815

         d. 0.3799

Q16. At a casino, gamblers are informed that chances of winning and losing a game have equal

         chances (50% each). The information is translated as the hypothesis H₀: π = 0.5 vs H₁: π ≠ 0.5,

          where π denotes the population proportion of gamblers who win the game. For its verification,

          on a randomly selected day 400 gamblers played and 228 won the game. At α=0.05, the test (s)

          statistic (s) that can used to verify H₀:

          a. Only χ²not Z

          b. Only Z and not χ²

     c. Both Z and χ²         

          d. Only t and neither Z nor χ²

Q17. To avoid panic, it is advertised by NYC authority that only 4.5% of health workers are

          infected while on duty for COVID-19. To estimate the advertised proportion, a random sample

          of 150 health workers was selected and 3 of them were tested positive. Confidence interval (CI)

          estimate using CI=p ± Z_α/2 √p(1-p)n was calculated. Choosing α (alpha) = 0.01, underscore the

          correct statement (s) from the following:

          a. CI estimates and Z_STAT to test hypothesis H₀: π=0.045 vs π ≠ 0.045 give same decision                

          b. π=0.045 can be used for calculating best CI estimates

         c. CI provides more than 5% estimates of infected workers        

         d. insufficient data to provide evidence to support NYC advertisement

Q18. Health Department, NYC plans to release a new drug commercially that is expected to lower

         blood pressure effectively. The drug was administered to a random sample of 20 patients.

         Among these patients the decrease in the blood pressure of 7 patients was recorded and no

         change in blood pressure of other patients was recorded . The decrease in the blood pressure of

         the 7 patients was 21, 22, 16, 19, 20, 23, 19. At 1% level of significance, to support the

         release of drug, Health Department used

         a. mean decrease in blood pressure =20 for test of hypothesis

         b. Critical value t_α/2 = 3.707 was valid value to compare t_STAT

         c. Degree of freedom = 6 were used to get t_α/2 for the conclusion                     

        d none of the above a, b, c are correct for conclusion       

Q19. A major dish network chain is considering opening a new office in an area that currently does

         not have any office to serve residents of that area. The chain will open store only if more than

         7,200 of the 24,000 households in the area shows interest to get dish network installation in their

         houses. A telephone poll of 625 randomly selected households in the area shows that 425

         households are not interested in the dish network installations. Using 95% confidence level, the

         chain should

          a. use sample proportion 0.68 for making decision to open the office  

          b. not open its office in the area        

        c. use Z_α=1.96 to make decision to open the office     

          d. consider all of a, b, c suggestions for making a decision to open office

Q20. The variability in the sample (denoted by SS) refers to sum of squares of deviations of values

         from their sample mean. The variability in the samples of gas consumption (miles per gallon) by

         Japanese and USA cars are respectively SS_JPN = 51 and SS_USA = 53 and population means of gas

         consumption are denoted by Ս_JPN and Ս_USA. The pooled variance of these variabilities denoted

         by S_p² = 4. Then H₀: Ս_JPN = Ս_USA vs H₁: Ս_JPN ≠ Ս_USA can be tested by t test. The calculated value

         of t_STAT = 2.0. At α(alpha)=0.05, the true statement (s) is/are:

         a. H₀ is accepted

         b. t_α/2 can’t be found to make decision

         c. Degrees of freedom not known to get t_α/2

         d, Nothing can be underscored from a, b, c because of incomplete data

Answer 1

Q13. Option: d. H0 is assumed true under which pooled proportion will be unbiased for π.

Q14. Since P(ZSTAT>2.12)=P(ZSTAT<-2.12)=0.017 (since standard normal distribution is symmetric about zero) then p-value for two tail test=2P(ZSTAT>2.12)=2*0.017=0.034.

Since 0.02<P-value<0.04 so the null should be rejected at =0.04 and the null hypothesis fail to reject at =0.02. So

Option: c. the null hypothesis should be rejected at α=0.04.

Q15. Option: b. 0.3809.

Q16. Option: c. Both Z and χ2

Q17. Since np=150*0.045=6.75>5 and n(1-p)>5 so we can use normal approximation and 99% CI is

(0.000000, 0.049444).

Option: a. CI estimates and ZSTAT to test hypothesis H0: π=0.045 vs π ≠ 0.045 give same decision

Q18. Option: d none of the above a, b, c are correct for conclusion  (Because it is one sample proportion test or we can use one sample sign test).