Question

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 41 subscribers to Plan A is $55,400 with a standard deviation of $8,200. This distribution is positively skewed; the coefficient of skewness is not larger. For a sample of 35 subscribers to Plan B, the mean income is $57,000 with a standard deviation of $9,600.

At the 0.02 significance level, is it reasonable to conclude the mean income of those selecting Plan B is larger?

a. State the decision rule. (Negative answer should be indicated by a minus sign. Round the final answer to 3 decimal places.)

Reject H₀ if t > .

b. Compute the value of the test statistic. (Negative answer should be indicated by a minus sign. Round the final answer to 3 decimal places.)

Value of the test statistic            

c. What is your decision regarding the null hypothesis?

(Click to select)  Reject  Do not reject    H₀. There is  (Click to select)  enough  not enough  evidence to conclude that the mean income of those selecting Plan B is  (Click to select)  not larger  larger  .

d. What is the p-value? (Round the final answer to 4 decimal places.)

Answer 1

Answer:

a)

Given,

n1 = 41

n2 = 35

degrees of freedom = n₁ + n₂ - 2

substitute n1 , n2 values

= 41 + 35 - 2

degrees of freedom = 74

The critical value can be given as follows,

i.e., one tailed test

P( t₇₄ < - 2.091 ) = 0.02

Here by observing this we can say that, the decision rule is given as we reject the null hypothesis Ho where the t < - 2.091

b)

To determine the test statistic

Consider the standard error which is given as follows,

SE = sqrt(s1^2/n1 + s2^2/n2)

substitute values

= sqrt(8200^2/41 + 9600^2/35)

= sqrt(4273142.857)

= 2067.1582

Now the test statistic value can be given as follows

t* = (X1bar - X2bar)/SE

substitute values

t* = (55400 - 57000) / 2067.1582

test statistic t* = - 0.7740

c)

To determine the decision regarding null hypothesis

Here, we see that the test statistic esteem lies in the non dismissal district in light of the fact that

i.e.,

- 0.7740 > - 2.091

therefore we can't dismiss the invalid speculation here.

The end here is that the mean salary of those choosing Plan B isn't not the same as the mean pay of those choosing Plan A.

d)

To determine the p value

p = P( t₇₄ < - 0.774 )

= 0.2207 [ since from the t table]

p value = 0.2207