Question

4. At a local high school, the juniors and seniors compete to see who has a better average on a statistics test. Ten randomly selected junior had an average of 38.0 points with a standard deviation of 2.9 and 14 randomly selected seniors had an average of 31.2 points with a standard deviation of 6.2. Assume both populations follow a normal distribution.

We want to perform a test to see if on average juniors perform better on the test.

You must do all of your calculations on the TI calculator.

DO NOT Use the Stats -> Tests for any of the problems.

a. Define the parameters and state the hypotheses. [3pts]

b. Find the observed sample statistic and the test statistic. Use proper notation for the sample statistic ( x or p ˆ ), and clearly identify your test statistic as either a z-score or a t-score. [4pts]

c. Find the p-value of your test statistic. You are required to graph the test statistic on its corresponding distribution, shade the appropriate area, and provide correct calculator documentation. Box final answer. [3pts]

d. Make a generic decision about H0 . [2pts]

e. Make a conclusion in the context of the problem. [2pts]

Answer 1

Let - population mean of statistics test for Juniors

11

- population mean of statistics test for seniors

112

n1 = sample size of juniors = 10

T1 = sample mean of juniors = 38

s₁ - sample standard deviation for juniors = 2.9

n2 = sample size of seniors = 14

T 2 sample mean of seniors = 31.2

s₂ - sample standard deviation of seniors = 6.2

Claim: Juniors perform better on test that is $\mu_1 > \mu_2$

a. Parameters.

- population mean of statistics test for Juniors

11

- population mean of statistics test for seniors

112

The null and alternative hypotheses are,

$H_0: \mu_1 \leq \mu_2\ and \ H_1: \mu_1 > \mu_2$

b. Observed sample statistics are means of both sample.

$\bar x_1 = 38$

$\bar x_2 = 31.2$

Here the population standard deviations are unknown so t test that is t score is used.

The formula of t test statistics

$t = \frac{\bar x_1 - \bar x_2}{\sqrt{s_{p}^{2}(\frac{1}{n_1}+\frac{1}{n_2})}}$

Where $s_{p}^{2} = pooled \ variance = \frac{(n_1-1)s_{1}^{2}+(n_2-1)s_{2}^{2}}{n_1+n_2-2}$

$s_{p}^{2} =\frac{(10-1)2.9^2+(14-1)6.2^2)}{10+14-2}=26.155$

$t = \frac{\bar x_1 - \bar x_2}{\sqrt{s_{p}^{2}(\frac{1}{n_1}+\frac{1}{n_2})}}=\frac{38-31.2}{\sqrt{26.155(\frac{1}{10}+\frac{1}{14})}} = 3.211$

Test statistics = 3.211

c) P-value

The alternative hypothesis contains the greater than sign, so the test is right tailed test.

Degrees of freedom = n1 + n2 - 2 = 10 + 14 - 2 = 22

P-value = P(t > test statistics) = P(t > 3.211)

Using the exacrl function =tdist(test statistics, df, tail) = tdist(3.211, 22, 1) = 0.0020

P-value = 0.0020

P-value = 0.0020

Rejection region P-value = 0.0020 t = 3.211

d. Decision about H0

Decision rule: If P-value > alpha then fail to reject the null hypothesis otherwise reject the null hypothesis.

Alpha is not given, take it as 0.05

P-value 0.0020 is not greater than 0.05 so reject the null hypothesis.

e. Conclusion:

Reject the null hypothesis, that is there is sufficient evidence to support the claim that juniors perform better on the test.