Questions: (a) State the null hypothesis (H0) and the alternative hypothesis (H1). (H0:) (H1:) (b) Determine the type of test statistic to use. (Choose one) (c) Find the value of the test statistic. (Round to three or more decimal places.) (d) Find the (p)-value. (Round to three or more decimal places.) (e) Can we conclude that the mean IQ score of this year's class is greater than that of previous years? Yes No Check

(a) State the null hypothesis (H0) and the alternative hypothesis (H1). (H0:) (H1:) (b) Determine the type of test statistic to use. (Choose one) (c) Find the value of the test statistic. (Round to three or more decimal places.) (d) Find the (p)-value. (Round to three or more decimal places.) (e) Can we conclude that the mean IQ score of this year's class is greater than that of previous years? Yes No Check
Transcript text: (a) State the null hypothesis $H_{0}$ and the alternative hypothesis $H_{1}$. \[ \begin{array}{l} H_{0}: \square \\ H_{1}: \square \end{array} \] (b) Determine the type of test statistic to use. (Choose one) (c) Find the value of the test statistic. (Round to three or more decimal places.) $\square$ (d) Find the $p$-value. (Round to three or more decimal places.) $\square$ (e) Can we conclude that the mean IQ score of this year's class is greater than that of previous years? Yes No Check
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Solution

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Solution Steps

Step 1: State the Hypotheses

The null hypothesis \( H_0 \) and the alternative hypothesis \( H_1 \) are stated as follows: \[ H_0: \text{The mean IQ score of this year's class is equal to that of previous years } (\mu = 100) \] \[ H_1: \text{The mean IQ score of this year's class is greater than that of previous years } (\mu > 100) \]

Step 2: Determine the Type of Test Statistic

The type of test statistic used for this hypothesis test is a \( t \)-test, as we are dealing with a sample mean and sample standard deviation.

Step 3: Calculate the Test Statistic

The sample mean \( \bar{x} \) is calculated as follows: \[ \bar{x} = \frac{\sum_{i=1}^N x_i}{N} = \frac{1325}{10} = 132.5 \]

The standard error \( SE \) is calculated using the formula: \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{15.138}{\sqrt{10}} \approx 4.787 \]

The test statistic \( t \) is then calculated as: \[ t = \frac{\bar{x} - \mu_0}{SE} = \frac{132.5 - 100}{4.787} \approx 6.789 \]

Step 4: Calculate the p-value

For a right-tailed test, the p-value is calculated as: \[ P = 1 - T(z) \approx 0.0 \]

Step 5: Conclusion

Since the p-value \( 0.0 \) is less than the significance level \( \alpha = 0.05 \), we reject the null hypothesis. Therefore, we can conclude that the mean IQ score of this year's class is greater than that of previous years.

Final Answer

\[ \boxed{H_0: \mu = 100, \quad H_1: \mu > 100, \quad \text{Test Statistic} = 6.789, \quad p\text{-value} = 0.0, \quad \text{Conclusion: Yes}} \]

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