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

(a) The null hypothesis
`\( H_0: p = 0.125 \)` suggests that James' strikeout rate using the new stance is equal to his historical rate of 12.5%. The alternative hypothesis
`\( H_a: p < 0.125 \)` indicates that James aims to reduce his strikeout rate, making it less than 12.5%.

(b) After conducting a one-proportion hypothesis test in Excel, the test statistic (z) is calculated to be approximately -0.973, and the p-value is about 0.163, rounded to three decimal places.

Step-by-step explanation:

James conducts a one-proportion hypothesis test to examine if his new batting stance has a significant effect on reducing his strikeout rate against left-handed pitchers. The null hypothesis
`\( H_0 \)` assumes that the proportion p of strikeouts remains at 12.5%, while the alternative hypothesis
`\( H_a \)` suggests that p is less than 12.5%, reflecting James' desire to improve.

To perform the test, James collects data from 200 simulated at-bats, resulting in 16 strikeouts. Using Excel or statistical software, the test statistic z is computed by comparing the observed proportion of strikeouts to the expected proportion under the null hypothesis. The negative value of z indicates that the observed proportion is less than expected, aligning with James' goal.

The p-value, the probability of obtaining results as extreme as observed under the null hypothesis, is calculated. In this case, the p-value of 0.163 is greater than the significance level of 0.05, suggesting that there is not enough evidence to reject the null hypothesis. James may not have successfully reduced his strikeout rate with the new batting stance based on the current data.

Answer 2

James is testing whether his new batting stance reduces his strikeout rate against left-handed pitchers using a one-proportion hypothesis test with a 5% significance level. The test requires calculating the test statistic and p-value using Excel, and then comparing the p-value to the significance level to determine if the null hypothesis can be rejected.

Step-by-step explanation:

James, a baseball player, is conducting a one-proportion hypothesis test to determine if his new batting stance reduces his strikeout rate against left-handed pitchers below his previous rate of 12.5%. Based on 200 simulated at-bats with a pitching machine, where he struck out 16 times, he wants to know if there is statistical evidence to suggest that the true proportion of strikeouts with his new stance is less than 12.5%.

Steps for Hypothesis Testing Using Excel:

Compare the p-value to the significance level, α = 0.05. If the p-value is less than α, we reject the null hypothesis.

The p-value helps us determine the strength of the evidence against the null hypothesis. A smaller p-value represents stronger evidence against H0. If the p-value is less than the significance level, it indicates that such a result is unlikely to occur due to random chance, given that H0 is true.

The exact values of the test statistic z and the p-value are not provided here, as these would be calculated using Excel based on the provided data (200 at-bats, 16 strikeouts).