Question

Let x be a random variable representing dividend yield of Australian bank stocks. We may assume that x has a normal distribution with σ =2.9%. A random sample of 14 Australian bank stocks has a sample mean of ?= 7%. For the entire Australian stock market, the mean dividend yield is µ = 5.2%. We are going to investigate if the dividend yield of all Australian bank stocks is higher than 5.2%? α= 0.05.

a) Find the test statistic.

b) Find the P-value.

c) Find critical value(s)(Cr. Val.).Use any method to either reject or fail to reject the claim

Answer 1

Given

A random sample of 14 Australian bank stocks has a sample mean of ?= 7% or 0.07.

We may assume that x has a normal distribution with σ =2.9% or 0.029

Thus = 0.07 ,    and σ = 0.029   { population standard deviation known }

For the entire Australian stock market, the mean dividend yield is µ = 5.2%. or 0.052

We are going to investigate if the dividend yield of all Australian bank stocks is higher than 5.2%? α= 0.05.

To test

H0 : µ = 0.052   ( dividend yield of all Australian bank stocks are not higher than 0.052 or 5.2% )

H0 : µ > 0.052    ( dividend yield of all Australian bank stocks is higher than 0.052 or 5.2% )

Note that - Since population standard deviation known we can use Z-score

a) Find the test statistic.

TS =

      = = 2.322408

Thus test statistic. = 2.322408

b)

Since alternative hypothesis is of " > " type so these is right tail test

Thus P-value = P ( Z > TS ) = P ( Z > 2.322408 )

where Z ~ N(0,1)

Now P ( Z > 2.322408 ) = 1 - P ( Z < 2.322408 )

P ( Z < 2.322408 ) can be computed from statistical book or more accurately from any software like R,Excel

From R

> 1-pnorm(2.322408 ,m=0,sd=1)          # 1 - P ( Z < 2.322408 )
[1] 0.01010549

Hence    P ( Z > 2.322408 ) = 1 - P ( Z < 2.322408 ) = 0.01010549

P-value = P ( Z > 2.322408 ) = 0.01010549

So P-value is 0.01010549

c) Find critical value(s)(Cr. Val.).Use any method to either reject or fail to reject the claim .

Since population standard deviation known we can use Z-score .

So at α= 0.05 critical value is 1.64 { for one tail test }

#1 Using Critical value

Since Test Statistics TS = 2.322408 > 1.64

we reject null hypothesis at 5% of level of significance

#2 Using P-value approach

Here P-value is 0.01010549   < 0.05

i.e P-value < α= 0.05

So we reject null hypothesis at 5% of level of significance

Conclusion -

Since we have rejected null hypothesis we conclude that dividend yield of all Australian bank stocks is higher than 5.2%