Question

Aron
	1	2	3	4	Average
8/6/2017	90	138	118	105	112.8
8/19/2017	162	101	120	145	132
9/16/2017	101	129	132	111	118.3
Average	117.7	122.7	123.3	120.3	121

Mjorgan
	1	2	3	4	Average
8/6/2017	115	88	94	102	99.8
8/19/2017	89	75	77	90	82.8
9/16/2017	74	110	117	90	97.8
Average	92.7	91	96	94	93.4

Mjorgan believes that there is no distinct difference within her first game played on each night, and her last. She does believe that there IS a difference between the three times she has bowled, though – as she improves. Run a statistical test to either confirm or refute Mjorgan’s analysis of the variation in her games.

Answer 1

Testing Hypothesis

In these example we want to see whether there is no distinct difference within her first game played on each night, and her last.

so our null hypothesis is, H0 :- there is no distinct difference within her first game played on each night, and her last. = d=0

where d is the difference between two means.

and alternative hypothesis :- there is distinct difference within her first game played on each night, and her last.=d0

so the test statistics is,

t =

we find the result of the problem by using r software as,

i) copy the given first and last game data as x and y resp. and paste in r as

=>x = c( first game)

x

=>y = c( last game)

=>y

=>result=t.test(x,y)

=>result

then we get , as follows, at 0.05 level of significance

the value of the test statistic t = -0.10559

the approximate p-value = 0.924

Decision rule :- i) if p value > 0.05 accept null hypothesis

ii) if p value < 0.05 reject null hypothesis

Result :- p value( 0.924) > 0.05 accept null hypothesis

conclusion :- yes. there is no distinct difference within her first game played on each night, and her last.

now we will see if there is any improvements between three times she has bowled.

null hypothesis =there is no any improvements between three times she has bowled.

alternative hypothesis = there is difference between three times she has bowled.

same procedure we will do, as above

we take result on r . first night result that she has bowled three times

=>y = c( first night 3 times bawl )

=>y

=>result=t.test(y)

=>result

then we get , as follows, at 0.05 level of significance

the value of the test statistic t = 12.095

the approximate p-value = 0.006767

Decision rule :- i) if p value > 0.05 accept null hypothesis

ii) if p value < 0.05 reject null hypothesis

Result :- p value( 0.006767) < 0.05 reject null hypothesis

conclusion :- there is difference between three times she has bowled.though she improves

Mjorgan’s analysis of the variation in her games is confirmed as she is true