To test the claim that the mean number of defined words on a page is greater than 48.0, we will perform a hypothesis test using the following steps:
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State the hypotheses:
- Null Hypothesis (H0): \(\mu \leq 48.0\) (the mean number of defined words is less than or equal to 48.0)
- Alternative Hypothesis (H1): \(\mu > 48.0\) (the mean number of defined words is greater than 48.0)
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Determine the significance level:
- The significance level is \(\alpha = 0.05\).
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Calculate the test statistic: We will use a one-sample t-test for the sample mean. The test statistic is calculated using the formula: \[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \] where:
- \(\bar{x} = 53.5\) (sample mean)
- \(\mu_0 = 48.0\) (hypothesized population mean)
- \(s = 15.7\) (sample standard deviation)
- \(n = 10\) (sample size)
Let's calculate the test statistic: \[ t = \frac{53.5 - 48.0}{15.7 / \sqrt{10}} = \frac{5.5}{15.7 / 3.1623} \approx \frac{5.5}{4.964} \approx 1.11 \]
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Determine the degrees of freedom: The degrees of freedom (df) for this test is \(n - 1 = 10 - 1 = 9\).
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Find the critical value: We will look up the critical value for a one-tailed t-test at \(\alpha = 0.05\) with 9 degrees of freedom. The critical value (from a t-distribution table) is approximately \(t_{0.05, 9} \approx 1.833\).
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Make the decision:
- If the calculated \(t\)-statistic is greater than the critical value, we reject the null hypothesis.
- If it is less than or equal to the critical value, we fail to reject the null hypothesis.
Our calculated \(t\)-statistic of \(1.11\) is less than the critical value of \(1.833\).
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Conclusion: Since \(1.11 < 1.833\), we fail to reject the null hypothesis.
Therefore, the correct choice is: A) Fail to reject the null hypothesis; there is not sufficient evidence to support the claim that the mean number of defined words on a page is greater than 48.0.