Question

Test the claim that LeBron James and Steph Curry average the same amount of points per game, on average.

Be sure to include the claim, critical values, test statistic and the conclusion.

For 70 games Lebron James averaged 30.1 points with a standard deviation of 9.112

For 70 games Steph Curry averaged 25.5 points with a standard deviation of 12.602

Answer 1

When the variances are known and the sample size is high, a z-test is used to assess if two population means vary. The mean of the two players is not the same.

What is a Z-test?

When the variances are known and the sample size is high, a z-test is used to assess if two population means vary.

In order to execute an appropriate z-test, the test statistic is expected to have a normal distribution, and nuisance factors such as standard deviation should be known.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

$\begin{array}{ccl} H_0: \mu_1 & = & \mu_2 \\\\ H_a: \mu_1 & \ne & \mu_2 \end{array}$

This corresponds to a two-tailed test, and a z-test for two means, with known population standard deviations, will be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the critical value for a two-tailed test is $z_c = 1.96$

The rejection region for this two-tailed test is R={z:∣z∣>1.96}

(3) Test Statistics

The z-statistic is computed as follows:

$\begin{array}{ccl} z & = & \displaystyle \frac{\bar X_1 - \bar X_2}{\sqrt{ {\sigma_1^2/n_1} + {\sigma_2^2/n_2} }} \\\\ & = & \displaystyle \frac{ 30.1 - 25.5}{\sqrt{ {9.112^2/70} + {12.602^2/70} }} \\\\ & = & 2.475 \end{array}$

$\begin{array}{ccl} z & = & \displaystyle \frac{\bar X_1 - \bar X_2}{\sqrt{ {\sigma_1^2/n_1} + {\sigma_2^2/n_2} }} \\\\ & = & \displaystyle \frac{ 30.1 - 25.5}{\sqrt{ {9.112^2/70} + {12.602^2/70} }} \\\\ & = & 2.475 \end{array}$

(4.) The decision about the null hypothesis

Since it is observed that |z| = 2.475 > z_c = 1.96, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p = 0.0133p=0.0133, and since p = 0.0133 < 0.05p=0.0133<0.05, it is concluded that the null hypothesis is rejected.

(5.) As it is concluded that the null hypothesis H₀ is rejected. Therefore, there is enough evidence to claim that the population mean μ1  is different than μ2.

Hence, the mean of the two players is not the same.

Learn more about Z-test:

https://brainly.com/question/15683598

#SPJ1