A study1 of 138 penalty shots in World Cup Finals games between 1982 and 1994 found that the goalkeeper correctly guessed the direction of the kick only 41% of the time. The article notes that this is ‘‘slightly worse than random chance."" We use these data as a sample of all World Cup penalty shots ever. Test at a 5% significance level to see whether there is evidence that the percent guessed correctly is less than 50%. The sample size is large enough to use the normal distribution. The standard error from a randomization distribution under the null hypothesis is SE=0.043.

So the p value obtained was a very low value and using the significance level given we have so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of interest is significantly lower than 0.5.

Step-by-step explanation:

Data given and notation

n=138 represent the random sample taken

estimated proportion of interest

is the value that we want to test

represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportions is lower than 0.5 or 505.:  

Null hypothesis:  

Alternative hypothesis:  

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

(1)  

The One-Sample Proportion Test is used to assess whether a population proportion is significantly different from a hypothesized value .

Calculate the statistic  

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

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 provided . The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

So the p value obtained was a very low value and using the significance level given we have so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of interest is significantly lower than 0.5.

let f (x ) is a function and c be a positive real number,   the graph of the function g (x ) = f (cx ) is represented as follows.

when, c > 1,   the function compresses it in the x - direction.

when, 0 < c < 1, the function stretches it.

the function is g ( x ) = (9x )2.

the heighest power of x in g ( x ) = (9x )2 is 2, so the parent function is square function.

the equation of parent function is f ( x ) = x 2 .

the function is g ( x ) = (9x )2 and the parent function is f ( x ) = x 2 .

so, the functions f and g have the following relationship.

g ( x ) = f ( 9x ).

compare the function g ( x ) = f ( 9x ) with g (x ) = f (cx ).

c = 9 ( > 1 ).

so, the function g (x ) = (9x )2 compresses it in the x - direction.

step-by-step explanation: