# HYPOTHESIS FOR BRUCE VS. BRETT

1 .PUBLISHED IN NEW ZEALAND

Here the null hypothesis is(Ho): Australian sample mean score is equal to the value shown on the website is 55

Alternative hypothesis(H1): Australian sample mean score is greater than to the value shown on the website is 55

Here population mean (µ) =55

Sample mean (Xbar) = 60

Sample standard deviation(s) =10

Assuming that the Australian scores are normally distributed and we check the hypothesis by the formula-

Z=(Xbar-µ)/(s/)

Then Z=(60-55)/(10/)

Z=1.5811

At the 5% level of significance the table value of right tail test is 1.645 and the calculated value is less than the table value. So we accept the alternative hypothesis and reject the null hypothesis. That means Australian mean score is greater than 55.

2. WINNERS ARE GRINNERS

Null hypothesis (Ho): New Zealand win 70% of the competition

Alternative Hypothesis (H1): New Zealand’s win percentage not equal to 70%

To test the hypothesis we use the formula-

Z=

Here,

P= population proportion of win is 0.70 from website result

p= sample proportion of win by random selection of game involving New Zealand that is 0.73

Q= (1-P)= population proportion of loss is 0.30 from website result

n = sample size =1000

Then , Z

Z=2.07

The tabulated value of Z at 10 % level of significance in two tailed test is 1.645 and the calculated value 2.07 is greater than 1.645, so we reject the alternative hypothesis and accept the null hypothesis means New Zealand wins 70% of the competition and the evidence is true.

3.BRUCE VS. BRETT

Null hypothesis (Ho): Brett’s expected score is equal to the Bruce’s expected score

Alternative hypothesis (H1): Brett’s expected score is greater than Bruce’s expected score

we use t distribution to test this-

t=

S2=(n1s12+n2s22)

X1bar=Brett’s sample mean

X2bar=Bruce’s sample mean

S=population standard deviation

s1= sample standard deviation of Brett’s group

s2= sample standard deviation of Bruce’s group

S2=137.714

t=0.5689

From the table value 14 d.f. 5% level of significance calculated t is less than tabulated t, from this we conclude that null hypothesis is rejected and alternative hypothesis is accepted means Brett’s expected score is greater than Bruce’s expected score.

4.

Null hypothesis: players are same

Alternative hypothesis: players are different

t=

S2=(n1s12+n2s22)=121.875

X1bar=Brett’s sample mean=33.4

X2bar=Bruce’s sample mean=32.4

S=population standard deviation

s1= sample standard deviation of Brett’s group=8.66

s2= sample standard deviation of Bruce’s group=10.965

t=0.143

from the tabulated value we see that calculated t is less than tabulated t at 1% and 5% level of significance. So, we conclude that one player is better than one.

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