Question 1196535: A marketing research company is estimating which of two soft drinks college students prefer. A random sample of 100 college students produced the following 95% confidence interval for the proportion of college students who prefer drink A: (0.262, 0.622). What would happen to the confidence interval if the sample size were changed to 1 000?
The interval would get narrower.
It is impossible to tell until the interval is constructed.
The interval would get wider.
There would be no change in the width of the interva
Answer by math_tutor2020(3817)   (Show Source):
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Answer: A) The interval would get narrower.
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Reason:
When dealing with proportions, the margin of error formula is
E = z*sqrt(phat*(1-phat)/n)
where,
z = critical value based on the confidence level
phat = sample proportion
n = sample size
If we fix z and phat to be constants, and let n vary, then E will decrease as n increases.
n goes up ---> E goes down
and vice versa
The variables move in opposite directions of one another.
The margin of error getting smaller is an indication that we're narrowing in on the true population proportion.
This is to be expected: The larger the sample, the better a sense we have about the true population proportion.
There is less guesswork on what the population proportion is.
Think of it like fishing for an elusive fish.
The more information you know, the smaller the net is needed to catch it.
The less info you know, the net will have to be larger.
The fishing net represents the confidence interval.
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Let's look at a numeric example.
I'll select a 95% confidence level and have phat = 0.50
Feel free to use whatever phat value you prefer, as long as 0 < phat < 1
Whatever you select, have it be fixed to a constant.
95% confidence means z = 1.96 approximately by use of a normal distribution table (aka Z table).
Such tables are in the appendix section of your stats textbook.
Or you can use an online one such as this
https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf
It says t table, but look at the very bottom row and it shows various Z values. The value 1.960 is right above the 95% confidence label.
Now consider a sample of n = 100
Compute the margin of error
E = z*sqrt(phat*(1-phat)/n)
E = 1.96*sqrt(0.50*(1-0.50)/100)
E = 0.098
Recompute with a larger sample of n = 1000
Keep the other input values the same.
E = z*sqrt(phat*(1-phat)/n)
E = 1.96*sqrt(0.50*(1-0.50)/1000)
E = 0.031
The results for each value of E are approximate.
The margin of error has gone from 0.098 to 0.031, which is a decrease.
This is one example of the confidence interval shrinking as the sample size gets larger.
Similar question:
https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.1196534.html
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