Question

According to a study conducted by the California Division of Labor Research and Statistics, roofing is one of the most hazardous occupations. Of 3,514 worker injuries that caused absences for a full workday or shift after the injury, 35% were attributable to falls from high elevations on level surfaces and 14% to burns or scalds. Assume that 3,514 injuries can be regarded as a random sample from the population of all roofing injuries in California. Construct a 90% confidence interval for the proportion of all injuries that are due to falls. (Please show in Excel)

Answer 1

Single sample proportion test

The α level confidence interval or (1-α)100% confidence interval is given by the form:

p(1-p) a/21/P1-

where, p is the estimated sample proportion of injuries due to falls. Here, n=3514 and p=0.35

Two-Tailed Tests: ZaE = NORM.S.INV(α/2)

NORM.S.INV stands for the inverse of the standard normal distribution.

Here α = 0.05, so, we use the following , =NORM.S.INV(0.10/2)

which gives us result

1.645

Now, Using the following we obtain the confidence interval.

if A1 has your critical value then, use in any other cell the following

=0.35-A1*SQRT(0.35*(1-0.35)/3514) for lower bound

=0.35+A1*SQRT(0.35*(1-0.35)/3514) for upper bound,

so 90% confidence interval for sample proportion is given by [0.3367,0.3632]