Question

Football team members suspect the coin used for the coin toss at the beginning of their games is unfair. They believe it turns up tails more often than it should if it were fair. The coach of the team decides to flip the coin 100 times and count the number of tails. His trial results in 55 tails. He decides to carry out a significance test. What is the p-value he obtains and the general conclusion that can be made at a 95% significance level?

Answer 1

Answer:

The p-value of the test is 0.1587.

The coins if fair.

Step-by-step explanation:

It is suspected by the football team members that the coin used for the coin toss at the beginning of their games is unfair, i.e. the coin turns up tails more often than it should if it were fair.

To test this suspicion we can use a one-proportion z-test.

A fair coin has an equal chance of landing a heads or tails, i.e.

P (Heads) = P (Tails) =0.50

The hypothesis can be defined as:

H₀: The coin is fair, i.e. p = 0.50.

Hₐ: The coin is unfair and turns up tails more than heads, i.e. p > 0.50.

The information provided is:

n = 100

X = number of tails = 55

The sample proportion of tails is:

$\hat p=\frac{55}{100}=0.55$

Compute the test statistic as follows:

$z=\frac{\hat p-p}{\sqrt{p(1-p)}{n}}=\frac{0.55-0.50}{\sqrt{\frac{0.50(1-0.50)}{100}}}=1$

The test statistic value is 1.

The decision rule is:

If the p-value of the test is less than the significance level then the null hypothesis will be rejected.

Compute the p-value as follows:

$p-value=P(Z>z)\\=P(Z>1)\\=1-P(Z<1)\\=1-0.8413\\=0.1587$

*Use a z-table for the probability.

The p-value of the test is 0.1587.

The significance level of the test is, α = 0.05.

p-value = 0.1587 > α = 0.05

The null hypothesis will not be rejected.

Hence, concluding that the coins if fair.

Answer 2

Answer:

The p-value is 0.159. He should fail to reject the null.

Step-by-step explanation:

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