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Summary OPMT1130 ALL NOTES + Midterm/Final
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- OPMT1130
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- British Columbia Institute Of Technology (BCIT )
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OPMT 1197Business Statistics
Lecture 16: Estimating the Population Proportion
1. Historically, Crest Toothpaste has been the market leader with 38% of the market share.
(38% of people use Crest toothpaste). Crest recently began selling whitening toothpastes in the
hope of increasing their market share. This increase could be offset by fierce competition by
their main rival, Colgate. Crest wants to estimate their current market share so they asked 160
adults what brand of toothpaste they buy. 48 of the 160 said they use Crest toothpaste.
(a) What is the point estimate of the percentage of people who use Crest toothpaste?
(b) Construct a 95% confidence interval for the percentage of people who use Crest.
(c) What is the margin of error at the 95% confidence level?
(d) Does it look like Crest’s market share has changed?
(e) Crest decides to take a much larger sample so that the error on the estimate is no more
than 3%. How many more people would need to be surveyed so the estimated (sample)
proportion is within 3% of the true population proportion of people who use Crest?
(f) Crest wonders if Colgate’s popularity is because people prefer peppermint over spearmint.
Crest wants to estimate the percentage of people who prefer peppermint flavoured
toothpaste to spearmint flavoured. They want the error on the estimate to be no more than
3% with 95% confidence. How many people would need to be surveyed?
2. You want to estimate the percentage of BCIT students who smoke on a regular basis. You want
the error on the estimate to be at most 3% with 95% confidence. How many BCIT students
would need to be surveyed to achieve your goal?
Solutions:
1. (a) 30% (b) 22.9% ≤ p ≤ 37.1% or 22.9% − 37.1%
I am 95% confident that between 22.9% and 37.1% of people use Crest
(c) I am 95% sure that the error is no bigger than 7.1% (5% chance it exceeds 7.1%)
(d) The highest estimate is 37.1% < 38% → looks like their market share is declining.
(e) 737 more people (f) 1,068 people
3. (a) 19.6% ≤ p ≤ 24.4% or 19.6 − 24.4% (b) 19/20 = 95% confidence level
(c) 2.0%, not 2.4%. They used p = 0.50 to calculate the margin of error, not 0.22 4. 92%
Pg 1 of 8
, OPMT 1197
Business Statistics
Summary of Formulas for Lectures 14-16
• Parameter: numerical measure from the population
• Statistic: numerical measure from the sample
Population Sample Statistic
Parameter (Point Estimate)
Mean µ x
Standard deviation σ s
Proportion p (or π) p
Standard deviation of the σ x
= σ n
s n
sample means = Standard Error
σp= p(1− p) p(1− p)
Standard deviation of the
sample proportions = Standard Error n n
Confidence Interval = point estimate ± margin of error
Confidence Interval Sample Size
Mean ̅± � �
√
σ known
if n < 30, sample must come from a normal population =� � 2
̅± � �
√ • always round up
Mean
• degrees of freedom = – 1 • use Z when finding
σ unknown
• if n < 30, sample must come from a normal
population
x 2
p ± z⋅ p(1− p) where p= =� � ̅(1 − ̅)
Proportion
n n • always round up
• need ̅ 5 and (1 − ̅) ≥ 5
≥
• if no idea use ̅= 0.5
• always use the normal table for proportions
Pg 2 of 8
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