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.
Question
Answer:
Answer:z=-2.09Step-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:[tex]H_0: p=0.50\\H_a: p <0.50\\[/tex](left tailed test at 5% significance level)Standard error = 0.043Sample proportion p = 0.41p difference = [tex]0.41-0.50=-0.09[/tex]Test statistic Z = p diff/std error= [tex]\frac{-0.09}{0.043} \\=-2.09[/tex]Z = -2.09
solved
general
10 months ago
3235