What is the solution of the equation?
1 answer:
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Answer:
The interval is (.0817, .118).
Step-by-step explanation:
Let's make this interval! The easy way is to go to STAT->TEST->A, 1-prop z-int, but I will show the long way.
<u>1. Conditions</u>
First we must check the conditions.
<em>Randomization condition: </em>The sample is given to be random, so we can assume independence.
<em>Success/failure condition: </em>Both np>10 and nq>10 must be met.
- np = 1030(103/1030) = 103 ✓
- nq = 1030(927/1030) = 927 ✓
<em>10% condition: </em>The sample cannot be greater than 10% of the population.
- 10n < N, 10(1030) < 200000 ✓
All conditions are met, so we can use <u>a 1-proportion z-interval</u> to represent the data.
<u></u>
<u>2. Mechanics</u>
<u>Find </u><u>, the sample proportion.</u>
103/1030 = .1
, then, the probability of not p, is 1 - .1 = .9
<u>Find z* (z star), the critical z-value.</u>
You can do this on your calculator using 2nd->VARS->3 (or you can memorize it - it's 1.96. Winky face.) This is called the invNorm function, that takes the left side area under the normal curve. But what do we plug in to the invNorm function?
invNorm(.975) = 1.96
<u>Plug values into the equation.</u>
± z*
.1 ± 1.96
.1 ± 1.96(.00935)
.1 - .0183 = .0817
.1 + .0183 = .118
(.0817, .118)
<u>3. Conclusion</u>
Based on this sample, we are 95% confident that the proportion of people the company contacts who may buy something is between .0817 and .118. That isn't very likely, but maybe they'll make a little profit!