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Answer:
There is enough evidence to support the claim that the percent that the goalkeeper guessed correctly is less than 50%.
Step-by-step explanation:
We have to perform an hypothesis test on a proportion.
We have a sample of size n=138, a sample mean of p=0.41.
The standard error is SE=0.043.
We want to test the claim that the real proportion is below 50%.
Then, the null and alternative hypothesis are:
The significance level is α=0.05.
The z-statistic can be calculated as:
The P-value (left tail test) for this z-statistic is:
The P-value is smaller than the level of significance, so the effect is significant. The null hypothesis is rejected.
There is enough evidence to support the claim that the percent that the goalkeeper guessed correctly is less than 50%.