Question

Adrian hopes that his new training methods have improved his batting average. Before starting his new regimen, he was batting 0.250 in a random sample of 56 at bats. For a random sample of 25 at bats since changing his training techniques, his batting average is 0.440. Determine if there is sufficient evidence to say that his batting average has improved at the 0.02 level of significance. Let the results before starting the new regimen be Population 1 and let the results after the training be Population 2. Step 2 of 3 : Compute the value of the test statistic. Round your answer to two decimal places.

Answer 1

According to the manufacturer's claim, we build an hypothesis test, find the test statistic and the p-value relative to this test statistic, we have that:

• The value of the test statistic is z = 1.65.
• The p-value of the test is of 0.0495 > 0.02, which means that there is not sufficient evidence to say that his batting average has improved at the 0.02 level of significance.

As the test involves a comparison of samples, it involves subtraction of normal variables, and for this, we need to understand the central limit theorem and subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

Before:

Average of 0.250 in 56 at bats, so:

$p_B = 0.25$

$s_B = \sqrt{\frac{0.25*0.75}{56}} = 0.0579$

After:

Average of 0.44 in 25 at bats, so:

$p_A = 0.44$

$s_A = \sqrt{\frac{0.44*0.56}{25}} = 0.0993$

Test if there was improvement:

At the null hypothesis, we test if there was no improvement, that is, the subtraction of the proportions is 0:

$H_0: p_A - p_B = 0$

At the alternative hypothesis, we test if there was improvement, that is, the subtraction of the proportions is positive, so:

$H_1: p_A - p_B > 0$

The test statistic is:

$z = \frac{X - \mu}{s}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, and s is the standard error.

0 is tested at the null hypothesis:

This means that $\mu = 0$

From the samples:

$X = p_B - p_A = 0.44 - 0.25 = 0.19$

$s = \sqrt{s_A^2 + s_B^2} = \sqrt{0.0579^2 + 0.0993^2} = 0.1149$

Value of the test statistic:

$z = \frac{X - \mu}{s}$

$z = \frac{0.19 - 0}{0.1149}$

$z = 1.65$

P-value of the test and decision:

The p-value of the test is the probability of finding a difference of at least 0.19, which is 1 subtracted by the p-value of z = 1.65.

Looking at the z-table, z = 1.65 has a p-value of 0.9505.

1 - 0.9505 = 0.0495.

The p-value of the test is of 0.0495 > 0.02, which means that there is not sufficient evidence to say that his batting average has improved at the 0.02 level of significance.

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