To calculate the test statistic for the given sample, we can use the formula for the t-statistic since we don't know the population standard deviation and our sample size is less than 30 (but is acceptable as the sample size is larger than 30, making the t-distribution approach normal distribution):
\[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \]
where:
- \(\bar{x} = 77.5\) (sample mean),
- \(\mu_0 = 74.1\) (population mean under the null hypothesis),
- \(s = 6.9\) (sample standard deviation),
- \(n = 53\) (sample size).
First, we will calculate the standard error (SE):
\[ SE = \frac{s}{\sqrt{n}} = \frac{6.9}{\sqrt{53}}. \]
Calculating \(\sqrt{53}\):
\[ \sqrt{53} \approx 7.2801, \]
Thus:
\[ SE \approx \frac{6.9}{7.2801} \approx 0.9485. \]
Now we can calculate the t-statistic:
\[ t = \frac{77.5 - 74.1}{0.9485} \approx \frac{3.4}{0.9485} \approx 3.5784. \]
Finally, rounding to three decimal places:
\[ \text{test statistic} = 3.578. \]
Next, we need to calculate the p-value for the one-tailed t-test. We will use the degrees of freedom, which can be calculated as:
\[ df = n - 1 = 53 - 1 = 52. \]
To find the p-value, we can use a t-distribution table or software/calculator. The p-value for a t statistic of 3.578 with 52 degrees of freedom can be found using a t-table or a statistical software/package.
Using statistical software or a calculator, we determine the p-value corresponding to the test statistic of 3.578 with 52 degrees of freedom.
The p-value is approximately \(0.0005\) (using appropriate statistical software or calculators).
Thus, the value to four decimal places is:
\[ \text{p-value} \approx 0.0005. \]
Summary of the results:
- Test statistic: \(3.578\)
- p-value: \(0.0005\)
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