Question

A coin was flipped 54 times and came up heads 34 times. At the .10 level of significance, is the coin biased toward heads?

a-1) H0: π ≤ .50 versus H1: π > .50. Choose the appropriate decision rule at the .10 level of significance.

Reject H0 if z > 1.282
Reject H0 if z < 1.282
(a-2) Calculate the test statistic. (Carry out all intermediate calculations to at least 4 decimal places. Round your answer to 3 decimal places.)

Test statistic:

b-1) Find the p-value. (Round your answer to 4 decimal places.)

p-value:

Answer 1

a-1) Reject H0 if z > 1.282

a-2)
`z=\frac{0.630 -0.5}{\sqrt{(0.5(1-0.5))/(54)}}=1.911`

b-1)
`p_v =P(Z>1.911)=0.0280`

Explanation:

1) Data given and notation

n=54 represent the random sample taken

X=34 represent the number of heads in the sample

`\hat p=(34)/(54)=0.630` estimated proportion of heads in the sample

`p_o=0.5` is the value that we want to test

`\alpha=0.1` represent the significance level (no given)

z would represent the statistic (variable of interest)

`p_v` represent the p value (variable of interest)

p= true proportion of heads

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the proportion of heads is higher than 0.5 :

Null hypothesis:
`p\leq 0.5`

Alternative hypothesis:
`p>0.5`

We assume that the proportion follows a normal distribution.

When we conduct a proportion test we need to use the z statistic, and the is given by:

`z=\frac{\hat p -p_o}{\sqrt{(p_o (1-p_o))/(n)}}` (1)

The One-Sample Proportion Test is used to assess whether a population proportion
`\hat p` is significantly (different,higher or less) from a hypothesized value
`p_o`.

Check for the assumptions that he sample must satisfy in order to apply the test

a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.

b) The sample needs to be large enough

`np_o =54*0.5=27>10`

`n(1-p_o)=54*(1-0.5)=27>10`

Choose the appropriate decision rule at the .10 level of significance.

For this case we need to find a value on the normal standard distribution that accumulates 0.1 of the area on the right tail of the distribution, and for this case is
`z_(\alpha)=1.282`

So the appropiate rule is:

Reject H0 if z > 1.282

3) Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

`z=\frac{0.630 -0.5}{\sqrt{(0.5(1-0.5))/(54)}}=1.911`

4) Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level is not provided usually is
`\alpha=0.1`. The next step would be calculate the p value for this test.

Since is a one side right tailed test the p value would be:

`p_v =P(Z>1.911)=0.0280`

Based on the p value obtained and using the significance level given
`\alpha=0.1` we have
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 10% of significance the proportion of heads is significantly higher than 0.5.