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.

(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].

5 answers

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

p^=xn

where 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:

p^=5251302

Calculating this gives:

p^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.

(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].
.

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:

p^±zp^(1p^)n

Where:

  • p^ is the sample proportion
  • n is the sample size
  • z is the z-value corresponding to the desired confidence level (for a 99.9% confidence level, z3.291)

From the previous calculation, we found:

  • p^=0.403
  • n=1302

Step 1: Calculate the standard error (SE)

SE=p^(1p^)n=0.403(10.403)1302=0.403×0.5971302

Calculating this step-by-step:

  1. Calculate p^(1p^): 0.403×0.5970.240501

  2. Divide by n: 0.24050113020.000185

  3. Take the square root: SE0.0001850.0136

Step 2: Calculate the margin of error (ME)

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

ME=zSE3.2910.01360.0448

Step 3: Calculate the confidence interval

Now we can construct the confidence interval:

p^MEtop^+ME

So the interval is:

0.4030.0448to0.403+0.0448

Calculating this gives:

0.3582to0.4478

Step 4: Round the results

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

0.358to0.448

Final Result

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

CI: 0.358<p<0.448

(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.

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.