Strategic business planning is required for the successful transfer of control between generations in family-owned companies. However, a survey of 449 privately held family firms whose annual turnover exceeded one million dollars it was found that 105 had no strategic business plan. Assuming that simple random sampling was employed to determine the sample for the survey, estimate with 85% confidence the proportion of family-owned companies operating without strategic business plans. Give the lower limit only and report your answer correct to three decimal places.
Solution:
Confidence interval for Population Proportion is given as below:
Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)
Where, P is the sample proportion, Z is critical value, and n is sample size.
We are given
Sample size = n = 449
Number of successes = x = 105
Sample proportion = P = x/n = 105/449 = 0.233853007
Confidence level = 85%
Critical Z value = 1.4395
(by using z-table)
Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)
Confidence Interval = 0.233853007 ± 1.4395* sqrt(0.233853007*(1 – 0.233853007)/449)
Confidence Interval = 0.233853007 ± 1.4395* 0.0200
Confidence Interval = 0.233853007 ± 0.0288
Lower limit = 0.233853007 - 0.0288 = 0.205
Upper limit = 0.233853007 + 0.0288 = 0.263
Lower limit = 0.205
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