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The manager of Lawn and Garden Services would like to estimate the proportion of employees' times spent performing various gardening or lawn care activities. The manager has made 400 random observations of a typical worker, with the following results: ACTIVITY NO. OF TIMES OBSERVEDMowing 200Trimming 80Raking 40Miscellaneous 80How confident can the manager be that the true proportion of time spent mowing is between .45 and .55?Multiple Choice99.7 percent90 percent99 percent95.44 percent95 percent

Question

The manager of Lawn and Garden Services would like to estimate the proportion of employees' times spent performing various gardening or lawn care activities. The manager has made 400 random observations of a typical worker, with the following results: ACTIVITY NO. OF TIMES OBSERVEDMowing 200Trimming 80Raking 40Miscellaneous 80How confident can the manager be that the true proportion of time spent mowing is between .45 and .55?Multiple Choice99.7 percent90 percent99 percent95.44 percent95 percent

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Solution

To answer this question, we need to calculate the confidence interval for the proportion of time spent mowing.

First, we calculate the observed proportion of time spent mowing. This is done by dividing the number of times mowing was observed by the total number of observations.

So, the observed proportion (p) = 200/400 = 0.5

Next, we calculate the standard error (SE) of the proportion. The formula for this is sqrt [ p(1 - p) / n ], where n is the total number of observations.

So, SE = sqrt [ 0.5(1 - 0.5) / 400 ] = 0.025

A confidence interval is calculated as p ± Z(SE), where Z is the Z-score corresponding to the desired confidence level.

The Z-scores for the given confidence levels are:

  • 99.7 percent: Z = 3
  • 90 percent: Z = 1.645
  • 99 percent: Z = 2.576
  • 95.44 percent: Z = 2
  • 95 percent: Z = 1.96

We need to find the confidence level where the interval (p - Z(SE), p + Z(SE)) includes the range (.45, .55).

After calculating the intervals for each confidence level, we find that the interval for 95.44 percent confidence level is (.45, .55).

So, the manager can be 95.44 percent confident that the true proportion of time spent mowing is between .45 and .55.

This problem has been solved

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