The General Social Survey asked 1302 people whether they performed any volunteer work during the past year. A total of 525 people said they did.

Question

(a) Find a point estimate for the proportion of people who performed volunteer work during the past year. Round the answer to at least three decimal places.

The point estimate for the proportion of people who performed volunteer work during the past year is [blank].

Answer 1

To find the point estimate for the proportion of people who performed volunteer work during the past year, you can use the formula:

$$\hat{p}=\frac{x}{n}$$

where $\hat{p}$ is the point estimate, $x$ is the number of people who performed volunteer work, and $n$ is the total number of people surveyed.

In this case:

• $x=525$
• $n=1302$

Now, plug in the values:

$$\hat{p}=\frac{525}{1302}$$

Calculating this gives:

$$\hat{p}\approx 0.403$$

So, rounding to at least three decimal places, the point estimate for the proportion of people who performed volunteer work during the past year is 0.403.

Answer 2

(b) Construct a 99.9% confidence interval for the proportion of people who performed volunteer work during the past year. Round the answer to at least three decimal places.

A 99.9% confidence interval for the proportion of people who performed volunteer work during the past year is [blank] < p < [blank].
.

Answer 3

To construct a 99.9% confidence interval for the proportion of people who performed volunteer work, we can use the formula for the confidence interval for a proportion:

$$\hat{p}\pm z^{\ast }\cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$

Where:

• $\hat{p}$ is the sample proportion
• $n$ is the sample size
• $z^{\ast }$ is the z-value corresponding to the desired confidence level (for a 99.9% confidence level, $z^{\ast }\approx 3.291$)

From the previous calculation, we found:

• $\hat{p}=0.403$
• $n=1302$

Step 1: Calculate the standard error (SE)

$$SE=\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=\sqrt{\frac{0.403(1-0.403)}{1302}}=\sqrt{\frac{0.403\times 0.597}{1302}}$$

Calculating this step-by-step:

1. Calculate $\hat{p}(1-\hat{p})$: $$0.403\times 0.597\approx 0.240501$$

2. Divide by $n$: $$\frac{0.240501}{1302}\approx 0.000185$$

3. Take the square root: $$SE\approx \sqrt{0.000185}\approx 0.0136$$

Step 2: Calculate the margin of error (ME)

Now we compute the margin of error using the z-value for 99.9% confidence:

$$ME=z^{\ast }\cdot SE\approx 3.291\cdot 0.0136\approx 0.0448$$

Step 3: Calculate the confidence interval

Now we can construct the confidence interval:

$$\hat{p}-ME\text{to}\hat{p}+ME$$

So the interval is:

$$0.403-0.0448\text{to}0.403+0.0448$$

Calculating this gives:

$$0.3582\text{to}0.4478$$

Step 4: Round the results

Rounding each end of the interval to at least three decimal places, we obtain:

$$0.358\text{to}0.448$$

Final Result

Thus, a 99.9% confidence interval for the proportion of people who performed volunteer work during the past year is:

$$\text{CI: }0.358<p<0.448$$

Answer 4

(c) A sociologist states that 49% of Americans perform volunteer work in a given year. Does the confidence interval contradict this statement? Explain.

The confidence interval [does or does not?] contradict the claim, because 0.49 [is or isnt?] contained in the confidence interval.

Answer 5

To determine whether the confidence interval contradicts the sociologist's statement, we need to analyze the value of 0.49 in relation to the confidence interval we calculated, which is approximately $(0.358,0.448)$.

• The confidence interval is: 0.358 < p < 0.448

• The sociologist's claim is that 49% (0.49) of Americans perform volunteer work in a given year.

Now, we check if 0.49 is contained within the confidence interval:

• Since $0.49$ is outside the interval $(0.358,0.448)$, it is not contained in the confidence interval.

Conclusion

Therefore:

The confidence interval does contradict the claim, because 0.49 isn't contained in the confidence interval.