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:

Respuesta :

Answer:

a-1) Reject H0 if z > 1.282

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

b-1) [tex]p_v =P(Z>1.911)=0.0280[/tex]  

Step-by-step explanation:

1) Data given and notation

n=54 represent the random sample taken  

X=34 represent the number of heads in the sample

[tex]\hat p=\frac{34}{54}=0.630[/tex] estimated proportion of heads in the sample

[tex]p_o=0.5[/tex] is the value that we want to test  

[tex]\alpha=0.1[/tex] represent the significance level (no given)  

z would represent the statistic (variable of interest)  

[tex]p_v[/tex] 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:[tex]p\leq 0.5[/tex]  

Alternative hypothesis:[tex]p>0.5[/tex]

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:  

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

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

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

[tex]np_o =54*0.5=27>10[/tex]

[tex]n(1-p_o)=54*(1-0.5)=27>10[/tex]

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 [tex]z_{\alpha}=1.282[/tex]

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:  

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

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 [tex]\alpha=0.1[/tex]. 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:  

[tex]p_v =P(Z>1.911)=0.0280[/tex]  

Based on the p value obtained and using the significance level given [tex]\alpha=0.1[/tex] we have [tex]p_v<\alpha[/tex] 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.