Homework answers / question archive / A viewer rating Research Company investigates the viewer rating of a TV drama

A viewer rating Research Company investigates the viewer rating of a TV drama

Statistics

Share With

---

A viewer rating Research Company investigates the viewer rating of a TV drama. The company found that 150 families watched the TV drama out of 1000 families randomly chosen. Let_Xi be a random variable such that 1f i-th family watched the TV drama, then X; = 1, and otherwise X; = 0. We suppose that the true viewer rating p 1s equal to the population mean of Xi, namely, p = E(X1) . Let P be the sample mean of Xi.

(1) (5 points) what is the value of??

(2) (5 points) What is the population variance of Xi ?

(A) logp. (B)p³. (C)p(l—3p), (D)p². (E)pU-p). (F)p’-p, (G)2p. (A) 3p,

(1) p(1 — 2p), (J) (A) — (PD) are all not correct.

(3) (5 points) What is the variance of P?

(A) np, (B)p²/n, (C) pU-np), (D)np³, (E)p0—p)/n, (F)n(p*-p), (G) 2pirr,

(H) pl —p)/n2, U)p(l-2p)/n, (J) (A) — (DP are all not correct, where n = 1000 in (A) — (1).

(4) (6 points) Let Tⁿ, be a statistic obtained by standardizing P Which form does T_n, take?

T,=(A) P22. (py YU  B=P) cy PPL. (py) VBE) Nalb=v)

Jp = p) J pl ~ p) Jp = np) p(l— p) J p(1 - 2p)

D D Vnp p-p

(F) per ) # Ff (H) , , (J)(A)-(D are all not correct,

p(l- 2p) Jp(l— p) p(l- p) p(l~ p)

where 7 = 1000 when n apperas in (A) — (I).

(5) (6 points) Conduct the hypothesis testing for the null hypothesis Ho: p = 0.1 against the alternative hypothesis H): p # 0.1 with the significance level a = 0.10, 0.05, and 0.01, using the test statistic T, obtained by substituting the null hypothesis into the answer in (4). What are the results? You can assume that n =1000 is large enough for the central limit theorem to work.

Results corresponding to a = (0.10, 0.05, 0.01) =

(A) (R, R,R), (B)(R,R,UR), (C)(R, UR, R), (D)(R, UR, UR), (E) CUR, R, R),

(F) (UR, R, UR), (G) (UR, UR, R), (CH) (UR, UR, UR), (1) (A) —(H) are all not correct,

Where R and UR imply “reject” and “not reject’, respectively.

A drug-manufacturing company developed a new medicine for cancers, and gave the new medicine to 400 cancer patients. Let X; be the change of the size of cancer of i-th patient after the administration. Let 4, denote the population mean of X;, namely, E(X;) = uw. The sample mean of Xj, 1 =1,...,100, was — 10 mm (millimeter). The sample standard deviation was 10 mm. Let o* denote the population variance of X;.Let” denote the sample mean. (6) (6 points) What is the variance of the sample mean”?

 

(A) jo (B) o?, (C) 4000°, (D) 4000. (E) 200. (F) 200°. (G) cv oOo” 2 o o

j00" (I) 0" (J) (A) — (J) are all not correct.

(7) (6 points) Let Z,, be a statistic obtained by standardizing /. Which form does Z,, take? (H)

Z, ~ (A) HOM (B) Vinit— 10 (C) He (D) Vie He (E) vidi (F) nfl — My Oo Oo oO oO o os!

G) A221 gp 20H ) Vino ©) Jo 20 Where 7 = 1000 when #7 apperas 1n (A) — (1). Aa-0.1 (1) (J) (A) — (dD) are all not correct,

(8) (6 points) Conduct the hypothesis testing for Ho: « = O against Hi: uw < 0 at significance level a= (0.1, 0.05, and 0.01 with a test statistic constructed by substituting appropriate values into the correct answer for Z, in (7). You can assume that the central limit theorem holds even when unknow parameter(s) in the standardized value is(are) replaced with their consistent estimator(s) (and this is actually true).

 Results corresponding to a = (0.10, 0.05, 0.01) =

(A) (R, R, RR), (B)(R,R,UR), (C)(R, UR, R), (D)(R, UR, UR), (E) (UR, R, R),

(F) (UR, R, UR), (G) (UR, UR, R), (CH) (UR, UR, UR), (1) (A) -—(H) are all not correct,

 where R and UR implies “reject” and “not reject’, respectively.

[For questions (9) — (12)] Suppose that we conduct a hypothesis testing with a test statistic Th. The test statistic T, follows U(O, 3), namely, the (continuous) uniform distribution with the interval (O, 3) under the null hypothesis. Set the significance level at 4%. Suppose that we reject the null hypothesis when 7;,1s greater than a critical value c. We know that T;, follows U

(1.5, 4.5) when the alternative hypothesis is true.

(9) (5 points) What is the probablity density function of the uniform the distribution?

(A) 0.5, (B) 1/3, (C) 2u, (D)u, (BE)1, (P)u/3, (G)0.5u, (H) 3/4, (1) 0.3,

(J) (A) — CD) are all not correct, (the range of support is accordingly defined)

(10) (6 points) what value of c should we use?

(A) 1.5, (B) 1.3, (C) 2.88, (D) 3.32, (E) 1.96, (F) 2.58, (G) 1.68, (H) 2.96, (1)0.12, (J) (A)—() are all not correct.

(11) (6 points) What is the power of the test againt the alternative considered in the problem?

(12) (6 points) What is the probability that Type II error occurs?

(13) (7 points) Suppose that under the same alternative as considered in the problem, the test statistics V, follows U(3 — d, 3 + d), where d >0. Consider the test that recjects the null hypothesis when V,, 1s greater than c, where c is the same value obtained in (10). For this test to be more powerfull than the test with T,, what condition does d have to satisfy?

(A)d>0.05, (B)d<0.05, (C) d1/4, (E)d=0.5,

 

(F)d>0.5, (G)d<0.5, (H)d*<09, (1)d<2/3, (J) (A)-() are all not correct.

 

[For questions (14) — (17)] We have n observations (a sample of size n). Consider about a test statistic Z, calculated from the n observations for a null hypothesis. We know that Z, follows the standard normal distribution under the null hypothesis. Furthermore, we also know that the test statistic follows M(1, 1) when an alternative hypothesis is true.

(14) (5 points) Suppose that we reject the null hypothesis when |Z,| > c. If we conduct the hypothesis testing at 5% significance level, what is the value of c?

(15) (7 points) When we conduct the hypothesis testing in (14), what is the power of the test against the alternative hypothesis mentioned in the problem?

(16) (6 points) Now, we have a different test statistic T_n We know that T;, follows N(d,, 1) when the alternative hypothesis is ture. Which condition should be satisfied for the test with T;, to have higher power than the test with Z, ?

(A) dn> 0.85, (B) dn< 0.05, (C) dn< 0.3, (D) dn> 1, (E) dn= 0.15, (F) di > 0.5, (G) dn< 1, CH) dn< 0.2, (I) dn< 1.5, (J) (A) — CD are all not correct.

(17) (6 points) Which condition should be satisfied as n goes to infinity for the test with 7), defined in (16) to be consistent?

(A) dn > 100, (B)dn<15, (C) dr—-0, (D)d,— 100, (E)!dn|— 100, (F)|dn lmao, (G)dn>1l, (H)dn

[Extra Questions: (18) — (20)] (These are extra questions that would require some additional knowlege beyond the level of this class. It is no problem if you cannot solve the problem).

Suppose that an 1.1.d. sequence X;, i=1,...,.2 follows a symmetric continuous distribution with mean Wo. Consider the hypothesis testing for Ho: 6o= O against the alternative H1: o> 0. We use two test Statistics:

X ly T, = vn and § =2Vn + DILX, > 0} - 05 ,

Oo WV ojey

Where Y=n' So" ¥,. G=n' DO" (¥,- XY and /{X; > 0} = 1 when X; > 0 and I{X; < 0}= 0.

It can be shown that the both test statistics coverge in distribtuion to the standard normal distribution under the null hypothesis, and both diverge to positive infinity under the alternative hypothesis, and thus both have the same rejection region in a positive area. Now, consider the alternative hypothesis H;™: 09= 6/J/n instead of H; (this kind of alternative is called a local alternative)

(18) (10 points) Under this alternative hypothesis, 7, coverges in distribution to N(a(6,o), 1) as n—o0, where a(@, o) is a function of 8 and o. What is the form of a(6@,c) ?

(A) £. (B) 6°, ©) eo, (D) 2 (E) 20logo, (F) 200°, (G) =

>

 

(H) ee (1) g (J) (A) — (D) are all not correct.