Let's break down the hypothesis test and solve the different parts step by step:

Given Information:

• Sample size $n = 1818$
• Sample mean $\overline{x} = 70.9$
• Sample standard deviation $s = 7.8$
• Significance level $\alpha = 0.01$
• Null hypothesis $H_0: \mu \leq 70$
• Alternative hypothesis $H_A: \mu > 70$

This is a one-tailed t-test because the alternative hypothesis suggests that $\mu$ is greater than 70.

Part a. State the decision rule.

The decision rule is based on the critical value from the t-distribution, which depends on the degrees of freedom and the significance level ($\alpha$).

• The degrees of freedom $df = n - 1 = 1818 - 1 = 1817$.
• The critical value $t_{\alpha, df}$ is determined from a t-table or statistical software. For a right-tailed test at $\alpha = 0.01$, we look for the critical value for $df = 1817$. Since degrees of freedom are quite large, we can approximate it using the standard normal distribution $Z$, but for precision, we use the t-distribution table.

So the decision rule will be:

• A. Reject the null hypothesis if the calculated value of the test statistic $t$ is greater than the critical value. Otherwise, do not reject.

Part b. State the calculated value of the test statistic.

The test statistic for a t-test is calculated using the following formula:

$t = \frac{\overline{x} - \mu_0}{\frac{s}{\sqrt{n}}}$

Where:

• $\overline{x} = 70.9$ (sample mean),
• $\mu_0 = 70$ (hypothesized population mean under $H_0$),
• $s = 7.8$ (sample standard deviation),
• $n = 1818$ (sample size).

Let's calculate the value of the t-statistic:

$t = \frac{70.9 - 70}{\frac{7.8}{\sqrt{1818}}}$

$t = \frac{0.9}{\frac{7.8}{42.643}} = \frac{0.9}{0.183} \approx 4.9180$

So the calculated value of the test statistic $t \approx 4.9180$.

Part c. State the conclusion.

Now, we compare the calculated test statistic $t = 4.9180$ with the critical value at $\alpha = 0.01$.

• From a t-table (or statistical software), for a one-tailed test with $df = 1817$ at $\alpha = 0.01$, the critical value $t_{\alpha, df}$ is approximately 2.326 (since it's close to the standard normal critical value for large $df$).

Since $t = 4.9180$ is greater than the critical value $2.326$, we reject the null hypothesis.

Conclusion:

We have enough evidence to reject the null hypothesis at the $\alpha = 0.01$ level. Therefore, we conclude that the population mean $\mu$ is significantly greater than 70.

Would you like further details or have any questions?

5 Related Questions:

1. How is the t-distribution different from the normal distribution in hypothesis testing?
2. What happens if the significance level $\alpha$ is changed to 0.05 instead of 0.01?
3. How do we determine whether to use a one-tailed or two-tailed test?
4. How would the decision change if the sample size were smaller, say $n = 30$?
5. What are Type I and Type II errors in hypothesis testing?

Tip:

When sample sizes are large (typically $n > 30$), the t-distribution approaches the normal distribution, allowing you to approximate critical values using the standard normal table for faster calculations.