Incorrect: Your answer is incorrect. (b) Use the results of part (a) and the appropriate computation formulas to compute the sample variance s2 and sample standard deviation s. s2

Let's break this down step-by-step for both part (a) and part (b).

Part (a): Calculation of Σx and Σx²

1. Sum of $x$ values ($\Sigma x$):

We already know from your input that: $\Sigma x = 22.6 + 32.1 + 57 + 66.5 + 70.2 + 89.1 = 337.5$

2. Sum of $x^2$ values ($\Sigma x^2$):

Now, we calculate $x^2$ for each $x$ value and sum them up.

• $(22.6)^2 = 510.76$
• $(32.1)^2 = 1032.41$
• $(57)^2 = 3249$
• $(66.5)^2 = 4422.25$
• $(70.2)^2 = 4928.04$
• $(89.1)^2 = 7932.81$

Now, sum these squared values: $\Sigma x^2 = 510.76 + 1032.41 + 3249 + 4422.25 + 4928.04 + 7932.81 = 23183.27$

So, $\Sigma x^2 = 23183.27$ (this confirms your previous calculation).

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Part (b): Sample Variance and Standard Deviation

Formula for sample variance $s^2$:

The formula for the sample variance is: $s^2 = \frac{\Sigma x^2 - \frac{(\Sigma x)^2}{n}}{n-1}$ Where:

• $\Sigma x^2 = 23183.27$
• $\Sigma x = 337.5$
• $n$ is the sample size. Here, $n = 6$.

Now, substitute the values into the formula: $s^2 = \frac{23183.27 - \frac{(337.5)^2}{6}}{6-1}$

First, calculate $\frac{(337.5)^2}{6}$: $(337.5)^2 = 113906.25$ $\frac{113906.25}{6} = 18984.375$

Now, subtract this from $\Sigma x^2$: $23183.27 - 18984.375 = 4198.895$

Now, divide by $n - 1$ (which is $6 - 1 = 5$): $s^2 = \frac{4198.895}{5} = 839.779$

Thus, the sample variance $s^2$ is approximately 839.779.

Formula for sample standard deviation $s$:

The sample standard deviation is simply the square root of the variance: $s = \sqrt{s^2} = \sqrt{839.779} \approx 28.98$

Final Answers:

• Sample variance ($s^2$) = 839.779
• Sample standard deviation ($s$) = 28.98

Would you like further clarification on any of the steps or have additional questions?

Here are 5 related questions to explore:

1. How does sample variance differ from population variance?
2. How would you calculate the population standard deviation for this data?
3. What is the significance of the degrees of freedom (n-1) in the variance formula?
4. How does the standard deviation help in interpreting data spread?
5. How would you handle outliers in a data set when calculating variance and standard deviation?

Tip: Always check that your squared values are correct and verify intermediate steps (such as dividing by $n$) to avoid calculation errors.