You wish to test the following claim (Ha) at a significance level of α=0.05.

Ho:μ=75.2
Ha:μ≠75.2

You believe the population is normally distributed, but you do not know the standard deviation. You obtain the following sample of data:

data

66

75.1

69.2

67

70.5

77.1

What is the test statistic for this sample? (Report answer accurate to three decimal places.)
test statistic =

What is the p-value for this sample? (Report answer accurate to four decimal places.)
p-value =

The p-value is...

less than (or equal to) α

greater than α

This test statistic leads to a decision to...

reject the null

accept the null

fail to reject the null

As such, the final conclusion is that...

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

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

The sample data support the claim that the population mean is not equal to 75.2.

There is not sufficient sample evidence to support the claim that the population mean is not equal to 75.2.

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.

---

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.

You wish to test the following claim (Ha) at a significance level of α=0.05. Ho:μ=75.2 Ha:μ≠75.2 You believe the population is normally distributed, but you do not know the standard deviation. You obtain the following sample of 66, 75.1, 69.2, 67, 70.5, 77.1. What is the test statistic for this sample? (Report answer accurate to three decimal places.) test statistic = What is the p-value for this sample? (Report answer accurate to four decimal places.) p-value = The p-value is... less than (or equal to) α greater than α This test statistic leads to a decision to... reject the null accept the null fail to reject the null As such, the final conclusion is that... There is sufficient evidence to warrant rejection of the claim that the population mean is not equal to 75.2. There is not sufficient evidence to warrant rejection of the claim that the population mean is not equal to 75.2. The sample data support the claim that the population mean is not equal to 75.2. There is not sufficient sample evidence to support the claim that the population mean is not equal to 75.2.

Explore the hypothesis test for the population mean with a sample of data 66, 75.1, 69.2, 67, 70.5, 77.1. This problem evaluates the significance level of α=0.05, calculating the test statistic and p-value while concluding evidence against the claim that the population mean is 75.2. Ideal for students in Grades 11-12.

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