Question

1. i flipped a coin 110 times and found that 63 were heads and 47 were tails. was there anything wrong with my coin or my flipping?

null hypothesis:
alternative hypothesis:

variables	o	e	o-e	(o-e)²	(o-e)²/e
					total

degrees of freedom: ______
p-value critical value: ______
calculated ( chi^2 ): ______
i (fail to reject / reject) my null hypothesis because the calculated chi square value of ____ is (lower than / higher than) the critical value of __ for ( p=0.05 ) at ____ degrees of freedom.

1. a poker dealing machine is supposed to deal cards randomly. in a test, you counted 1500 cards, and found the following:

spades: 380
hearts: 360
diamonds: 385
clubs: 375
null hypothesis:
alternative hypothesis:

variables	o	e	o-e	(o-e)²	(o-e)²/e
hearts
diamonds
clubs
					total:

degrees of freedom: ______
p-value critical value: ______
calculated ( chi^2 ): ______
(fail to reject / reject) my null hypothesis because the calculated chi square value of ____ is (lower than / higher than) the critical value of __ for ( p=0.05 ) at ____ degrees of freedom.

Explanation:

Answer:

Problem 1

Null hypothesis: The coin is fair (probability of heads = probability of tails = 0.5)
Alternative hypothesis: The coin is not fair (probability of heads ≠ probability of tails)

Variables	O	E	O - E	(O - E)²	(O - E)²/E
Tails	47	55	-8	64	1.1636
Total	110	110	0	128	2.3272

Degrees of freedom: \( 2 - 1 = 1 \)
P - Value critical value (for \( df = 1, \alpha = 0.05 \)): \( 3.841 \)
Calculated \( \chi^2 \): \( 2.3272 \)

I fail to reject my null hypothesis because the calculated chi square value of \( 2.3272 \) is lower than the critical value of \( 3.841 \) for \( p = 0.05 \) at \( 1 \) degrees of freedom.

Problem 2

Null hypothesis: The poker dealing machine deals cards randomly (probability of each suit is \( \frac{1}{4} \))
Alternative hypothesis: The poker dealing machine does not deal cards randomly (probability of at least one suit is different)

Variables	O	E	O - E	(O - E)²	(O - E)²/E
Hearts	360	375	-15	225	0.6
Diamonds	385	375	10	100	0.2667
Clubs	375	375	0	0	0
Total	1500	1500	0	350	0.9334

Degrees of freedom: \( 4 - 1 = 3 \)
P - Value critical value (for \( df = 3, \alpha = 0.05 \)): \( 7.815 \)
Calculated \( \chi^2 \): \( 0.9334 \)

I fail to reject my null hypothesis because the calculated chi square value of \( 0.9334 \) is lower than the critical value of \( 7.815 \) for \( p = 0.05 \) at \( 3 \) degrees of freedom.