Question

also what is Ho: and Ha:

would i use the p value and alpha value to come up with the conclusion or the alpha value and degrees of freedom from the x^2 table.

3. A gambler thinks that a die may be loaded, that is, the six numbers are not equally likely To test his suspicion, he rolled the die 150 times and obtained the following results: Number: 1 2 3 4 5 6 gume Frequency: 23 26 23 21 31 26 Do the data provide sufficient evidence that the die is loaded? Test at 10% level of significance.

Answer 1

Null hypothesis:Ho:All number on die appear with equal frequency.

Alternate  hypothesis:Ha:All number on die do not appear with equal frequency.

degree of freedom =categories-1=			5
for 0.1 level and 5 df :crtiical value X2 =				9.236
Decision rule: reject Ho if value of test statistic X2>9.236
applying chi square goodness of fit test:

	relative	observed	Expected	residual	Chi square
category	frequency(p)	Oi	Ei=total*p	R2i=(Oi-Ei)/√Ei	R2i=(Oi-Ei)2/Ei
1	1/6	23.000	25.000	-0.40	0.160
2	1/6	26.000	25.000	0.20	0.040
3	1/6	23.000	25.000	-0.40	0.160
4	1/6	21.000	25.000	-0.80	0.640
5	1/6	31.000	25.000	1.20	1.440
6	1/6	26.000	25.000	0.20	0.040
total	1.000	150	150		2.4800
test statistic X2 =		2.480

since test statistic does not falls in rejection region we fail to reject null hypothesis

we do not have have sufficient evidence to conclude that die is biased.