Question

A.

A cell phone company offers two plans to its subscribers. At the time new subscribers sign up, they are asked to provide some demographic information. The mean yearly income for a sample of 45 subscribers to Plan A is $45,200 with a standard deviation of $7,900. For a sample of 35 subscribers to Plan B, the mean income is $57,500 with a standard deviation of $6,400.

At the 0.1 significance level, is it reasonable to conclude the mean income of those selecting Plan B is larger? Hint: For the calculations, assume the Plan A as the first sample.

The test statistic is  . (Negative value should be indicated by a minus sign. Round your answer to 2 decimal places.)

The decision is (Click to select)RejectDo not reject the null hypothesis that the means are the same.

The p-value is (Click to select)greater than 0.10between 0.0005 and 0.005between 0.025 and 0.05between 0.01 and 0.025between 0.005 and 0.01between 0.05 and 0.10less than 0.0005.

B.

Suppose a manufacturer of Ibuprofen, a common headache remedy, recently developed a new formulation of the drug that is claimed to be more effective. To evaluate the new drug, a sample of 270 current users is asked to try it. After a one-month trial, 251 indicated the new drug was more effective in relieving a headache. At the same time, a sample of 390 current Ibuprofen users is given the current drug but told it is the new formulation. From this group, 352 said it was an improvement.

(1)	State the decision rule for .01 significance level: H0: πn ≤ πc; H1: πn > πc. (Round your answer to 2 decimal places.)

Reject H0 if z >

(2)	Compute the value of the test statistic. (Do not round the intermediate value. Round your answer to 2 decimal places.)

Value of the test statistic

(3)	Can we conclude that the new drug is more effective? Use the .01 significance level.

(Click to select)RejectDo not reject H0. We (Click to select)cancannot conclude that the new drug is more effective.

Answer 1

Part A

We have to use two sample t test for population means.

H₀: µ₁ = µ₂ versus H_a: µ₁ < µ₂

Test statistic formula for pooled variance t test is given as below:

t = (X1bar – X2bar) / sqrt[S_p²*((1/n1)+(1/n2))]

Where S_p² is pooled variance

S_p² = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

We are given

X1bar = 45200

S1 = 7900

n1 = 45

X2bar = 57500

S2 = 6400

n2 = 35

df = n1 + n2 – 2 = 45 + 35 – 2 = 78

α = 0.10

S_p² = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

S_p² = [(45 – 1)* 7900^2 + (35 – 1)* 6400^2]/(45 + 35 – 2)

S_p² = 53060000

t = (X1bar – X2bar) / sqrt[S_p²*((1/n1)+(1/n2))]

t = (45200 – 57500) / sqrt[53060000*((1/45)+(1/35))]

t = -12300/1641.6794

t = -7.4923

Test statistic = -7.49

P-value = 0.00

(by using t-table)

P-value < α = 0.10

P-value is less than 0.0005.

So, we reject the null hypothesis

Part B

(1)

Decision rule:

We are given α = 0.01, test is upper tailed. So, critical z value by using z-table or excel is given as below:

Critical z value = 2.3263

Reject H₀ if z > 2.33

(2)

Test statistic formula is given as below:

Z = (P1 – P2) / sqrt(P*(1 – P)*((1/N1) + (1/N2)))

Where,

X1 = 251

X2 = 352

N1 = 270

N2 = 390

P = (X1+X2)/(N1+N2) = (251 + 352)/(270 + 390) = 0.9136

P1 = X1/N1 = 251/270 = 0.92962963

P2 = X2/N2 = 352/390 = 0.902564103

Z = (0.92962963 – 0.902564103) / sqrt(0.9136*(1 – 0.9136)*((1/270) + (1/390)))

Z = 1.2170

Test statistic = Z = 1.22

(3)

By using above test statistic value, we have

P-value = 0.1118

(by using z-table)

P-value > α = 0.01

So, we do not reject the null hypothesis

We cannot conclude that the new drug is more effective.