The manager of Lawn and Garden Services would like to estimate the proportion of her employees' time spent performing various gardening and lawn care activities. She has made 400 random observations of a typical worker, with the following results: $$\begin{array} { l r }
\text { Activity } & \text { Times Observed } \
\hline \text { Mowing } & 200 \
\text { Trimming } & 80 \
\text { Raking } & 40 \
\text { Miscellaneous } & 80
\end{array}$$ Between what limits can the manager be 68.26 percent confident about the true proportion of time spent mowing?
A) 0 and .683
B) .475 and .525
C) .45 and .55
D) .425 and .575
E) .4 and .6

Question

The manager of Lawn and Garden Services would like to estimate the proportion of her employees' time spent performing various gardening and lawn care activities. She has made 400 random observations of a typical worker, with the following results: $$\begin{array} { l r } 
\text { Activity } & \text { Times Observed } \
\hline \text { Mowing } & 200 \
\text { Trimming } & 80 \
\text { Raking } & 40 \
\text { Miscellaneous } & 80
\end{array}$$ Between what limits can the manager be 68.26 percent confident about the true proportion of time spent mowing?
A) 0 and .683
B) .475 and .525
C) .45 and .55
D) .425 and .575
E) .4 and .6

Quizplus · Accepted Answer

To estimate the proportion of time spent mowing, we need to find the proportion of times mowing was observed out of the total observations.

Proportion of time spent mowing = 200/400 = 0.5

To find the confidence interval for this proportion, we will use the normal approximation to the binomial distribution, since we have a large sample size (n=400) and the proportion of success (mowing) is not too close to 0 or 1.

We will use the formula:

CI = p ± z*sqrt(p*(1-p)/n)

where p is the sample proportion, z* is the z-score corresponding to the desired confidence level (which is 68.26% in this case), and n is the sample size.

From the table of standard normal distribution, the z-score for a 68.26% confidence level is approximately 1.

Plugging in the values we get:

CI = 0.5 ± 1*sqrt(0.5*(1-0.5)/400)
CI = 0.5 ± 0.049
CI = (0.451, 0.549)

Therefore, we can be 68.26% confident that the true proportion of time spent mowing is between 0.451 and 0.549.

The only answer choice that contains these limits is B: 0.475 and 0.525.

The Manager of Lawn and Garden Services Would Like to Estimate