Questions: For the following conjecture, state the null and alternative hypothesis. Research indicates that the average amount of time elementary school children watch television is less than the average amount of time high schoolers watch television. Let μ1 represent the average amount of time elementary school children watch television. The null hypothesis is H0 : . The alternative hypothesis is H1 : μ1 μ2

For the following conjecture, state the null and alternative hypothesis. Research indicates that the average amount of time elementary school children watch television is less than the average amount of time high schoolers watch television. Let μ1 represent the average amount of time elementary school children watch television.

The null hypothesis is H0 : .

The alternative hypothesis is H1 :

μ1 μ2

Transcript text: For the following conjecture, state the null and alternative hypothesis. Research indicates that the average amount of time elementary school children watch television is less than the average amount of time high schoolers watch television. Let $\mu_{1}$ represent the average amount of time elementary school children watch television. The null hypothesis is $H_{0}$ : $\square$ . The alternative hypothesis is $H_{1}$ : $\square$ $\mu_{1}$ $\mu_{2}$

Solution

Solution Steps

Step 1: Define Hypotheses

We are testing the following hypotheses:

Null Hypothesis ($H_0$): The average amount of time elementary school children watch television is equal to or greater than that of high schoolers. \[ H_0: \mu_1 \geq \mu_2 \]
Alternative Hypothesis ($H_1$): The average amount of time elementary school children watch television is less than that of high schoolers. \[ H_1: \mu_1 < \mu_2 \]

Step 2: Calculate Standard Error

The Standard Error ($SE$) is calculated as follows: \[ SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{0.425}{5} + \frac{0.625}{5}} = 0.4583 \]

Step 3: Calculate Test Statistic

The test statistic ($t$) is computed using the formula: \[ t = \frac{\bar{x}_1 - \bar{x}_2}{SE} = \frac{2.4 - 5.0}{0.4583} = -5.6737 \]

Step 4: Calculate Degrees of Freedom

The degrees of freedom ($df$) are calculated using the formula: \[ df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}} = \frac{0.0441}{0.0057} = 7.7199 \]

Step 5: Calculate p-value

The p-value associated with the test statistic is: \[ P = T(t) = T(-5.6737) = 0.0003 \]

Step 6: Determine Critical Value

For a left-tailed test at a significance level of $\alpha = 0.05$, the critical value is: \[ \text{Critical value} = -1.8683 \]

Step 7: Conclusion

Since the calculated $t$-statistic of $-5.6737$ is less than the critical value of $-1.8683$ and the p-value of $0.0003$ is less than $\alpha = 0.05$, we reject the null hypothesis.