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, 26% 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 95% confidence interval for the proportion of all injuries that are due to falls. (PLEASE SHOW USING EXCEL FORMULAS)

a.Upper Bound: 0.2745, Lower Bound: 0.2455

b.Upper Bound: 0.3545, Lower Bound: 0.1855

c.Upper Bound: 0.2325, Lower Bound: 0.2955

d.Upper Bound: 0.4325, Lower Bound: 0.1255

Answer 1

For a single sample proportion test, an level confidence interval or (1-)100% confidence interval is given by the form:

where, is the estimated sample proportion of injuries due to falls. Here, n=3514 and =0.26

Two-Tailed Tests: = 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(0.05/2)

which gives us result

1.95996

Now, Using the following we obtain the confidence interval
if A1 has your critical value then, use in any other cell the following =0.26-A1*SQRT(0.26*(1-0.26)/3514) for lower bound and

=0.26+A1*SQRT(0.26*(1-0.26)/3514) for upper bound,

so 95% confidence interval for sample proportion is given by [0.2455,0.2745]

so (a) is the correct option.