Question

James is a baseball player who hits left handed. Based on his past statistics, his strikeout rate against left-handed pitchers is 12.5%. He would like to reduce this rate, so he changes his batting stance. To test whether it works, he uses a pitching machine to simulate 200 at bats. In these, he struck out 16 times.

James conducts a one-proportion hypothesis test at the 5% significance level, to test whether the true proportion of strikeouts against left-handed pitchers using James's new stance is less than 12.5%.

(a) H0:p=0.125; Ha:p<0.125, which is a left-tailed test.

(b) Use Excel to test whether the true proportion of strikeouts against left-handed pitchers using James's new stance is less than 12.5%. Identify the test statistic, z, and p-value from the Excel output, rounding to three decimal places.

Answer 1

We can model the number of strike outs using a binomial dist

This is because James simulates 200 bats (200 trials ) and strike outs are 16 (16 successes)

We are using to indicate sample proportion

We have to test whether true proportion is less than 12.5% = 0.125

Test

Test Statistic:

Where is the null hypothesis

Substituting the values

=

Test Stat = -2.3458

We will use a z-test to test true proportion and since we are ony testing for the left side this is a 1-tailed test.

p-value is the probability of the null hypothesis being true. If it less than our acceptance (level of significance), we reject it.

p - value =

= P (Z> | 2.346| )

= P( Z > 2.346)

we will use excel function '1 - normsdist(2.346)'. we use '1-' since we want greater than probability

p-value = 0.0095

since p-value < 0.05 (level of significance)

We reject the null hypothesis at 5% level of significance.

We conclude that the true proportion of strikeouts against left-handed pitchers using James's new stance is less than 12.5%.