Question

# Your friend tells you that the proportion of active Major League Baseball players who have a...

Your friend tells you that the proportion of active Major League Baseball players who have a batting average greater than .300 is different from 0.74, a claim you would like to test. The hypotheses here are Null Hypothesis: p = 0.74, Alternative Hypothesis: p ≠ 0.74. If you take a random sample of players and calculate p-value for your hypothesis test of 0.9623, what is the appropriate conclusion? Conclude at the 5% level of significance.

Question 15 options:

 1) We did not find enough evidence to say a significant difference exists between the proportion of MLB players that have an average higher than .300 and 0.74
 2) We did not find enough evidence to say the proportion of active MLB players that have an average higher than .300 is larger than 0.74.
 3) The proportion of active MLB players that have an average higher than .300 is equal to 0.74.
 4) The proportion of active MLB players that have an average higher than .300 is significantly different from 0.74.
 5) We did not find enough evidence to say the proportion of active MLB players that have an average higher than .300 is less than 0.74.

#### Homework Answers

Answer #1

Answer:

Given,

Test statistic, z = 0.9623

By observing we can say that It is a two tailed test.

Here it is 5% Significance level,

i.e.,

= 0.05

Now the P-value for a two tailed test =2*(1 - NORM.S.DIST(ABS(0.9623), TRUE)

P value = 0.3359

P - value = 0.3359 > 0.05 here it is greater than the significance level, so we fail to reject the null hypothesis Ho.

So we can conclude that Option C is correct answer.

i.e.,

The proportion of active MLB players that have an average higher than .300 is equal to 0.74.

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