Question

A study 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. Let p represent the proportion of all World Cup penalty kicks for which the goalkeeper correctly guesses the direction of the kick.
a. State hypotheses in terms of a single proportion p.

b. Write the value of the sample statistic, using correct notation. Use the sample statistic and the standard error provided to calculate a z-statistic. Show your calculation.

Answer 1

Answer:

z=-2.09

Step-by-step explanation:

Given that a study  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.

Set up hypotheses as:

$H_0: p=0.50\\H_a: p$

(left tailed test at 5% significance level)

Standard error = 0.043

Sample proportion p = 0.41

p difference = $0.41-0.50=-0.09$

Test statistic Z = p diff/std error

= $\frac{-0.09}{0.043} \\=-2.09$

Z = -2.09