To perform the hypothesis test, we first need to calculate the sample mean and sample standard deviation from the provided data. Then we can compute the test statistic and the p-value.

Given Data:

• Sample dataTo perform the hypothesis test, we first need to calculate the sample mean and sample standard deviation from the provided data. Then we can compute the test statistic and the p-value.

Given Data:

• Sample 66, 75.1, 69.2, 67, 70.5, 77.1

Step 1: Calculate Sample Mean ( $\bar{x}$ )

$\bar{x} = \frac{66 + 75.1 + 69.2 + 67 + 70.5 + 77.1}{6} = \frac{425.9}{6} \approx 70.9833$

Step 2: Calculate Sample Standard Deviation ( $s$ )

$s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}}$ Where $n$ is the sample size (6).

1. Calculate the squared deviations:

   • $(66 - 70.9833)^2 \approx 24.0314$
   • $(75.1 - 70.9833)^2 \approx 16.7068$
   • $(69.2 - 70.9833)^2 \approx 3.1628$
   • $(67 - 70.9833)^2 \approx 16.0914$
   • $(70.5 - 70.9833)^2 \approx 0.2339$
   • $(77.1 - 70.9833)^2 \approx 37.2490$

2. Sum of squared deviations: $24.0314 + 16.7068 + 3.1628 + 16.0914 + 0.2339 + 37.2490 \approx 97.4753$

3. Calculate $s$: $s = \sqrt{\frac{97.4753}{6 - 1}} \approx \sqrt{19.4951} \approx 4.4147$

Step 3: Calculate Test Statistic ( $t$ )

Using the formula for the t-test statistic: $t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}$ Where:

• $\mu_0 = 75.2$
• $n = 6$

$t = \frac{70.9833 - 75.2}{4.4147/\sqrt{6}} \approx \frac{-4.2167}{1.8049} \approx -2.337$

Step 4: Calculate the p-value

For a two-tailed t-test with $n - 1 = 5$ degrees of freedom:

Using a t-table or calculator, we find the p-value corresponding to $t \approx -2.337$ and 5 degrees of freedom.

Using a t-distribution calculator:

• p-value $\approx 0.0524$

Conclusion

1. Test Statistic: $t \approx -2.337$ (to three decimal places: -2.337)
2. p-value: $\approx 0.0524$ (to four decimal places: 0.0524)

Decision:

• The p-value (0.0524) is greater than $\alpha = 0.05$.
• Therefore, we fail to reject the null hypothesis.

Final Conclusion:

There is not sufficient evidence to warrant rejection of the claim that the population mean is not equal to 75.2.

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If you have any questions or need further details, feel free to ask! Here are some follow-up questions to consider:

1. How would the conclusion change if the significance level were set to 0.01?
2. What assumptions must be satisfied for this test to be valid?
3. How would increasing the sample size affect the test statistic and p-value?
4. What is the power of this test?
5. Can you explain the implications of failing to reject the null hypothesis in a practical context?

Tip: Always check the conditions for using the t-test, such as normality and independence of observations.