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bezimeni [28]
3 years ago
13

A simple random sample of 85 students is taken from a large university on the West Coast to estimate the proportion of students

whose parents bought a car for them when they left for college. When interviewed, 51 students in the sample responded that their parents bought them a car. What is a 95% confidence interval for p, the population proportion of students whose parents bought a car for them when they left for college
Mathematics
1 answer:
ahrayia [7]3 years ago
6 0

Answer:

95% confidence interval for p, the population proportion of students whose parents bought a car for them when they left for college

(0.4958 , 0.7041)

Step-by-step explanation:

<u><em>Step(i):-</em></u>

Given that the sample size 'n' = 85

The sample proportion

            p = \frac{x}{n} = \frac{51}{85} = 0.6

95% confidence interval for p, the population proportion of students whose parents bought a car for them when they left for college

(p^{-} -Z_{0.05} \sqrt{\frac{p(1-p)}{n} } , p^{-} +Z_{0.05} \sqrt{\frac{p(1-p)}{n} })

<u><em>Step(ii):-</em></u>

Given that the level of significance

α = 0.05

Z₀.₀₅ = 1.96

(0.6 -1.96 \sqrt{\frac{06(1-06)}{85} } , 0.6 +1.96 \sqrt{\frac{0.6(1-0.6)}{85} })

(0.6 - 0.104138 , 0.6 +0.104138)

(0.4958 , 0.7041)

<u><em>Final answer:-</em></u>

95% confidence interval for p, the population proportion of students whose parents bought a car for them when they left for college

(0.4958 , 0.7041)

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8 0
3 years ago
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kupik [55]

Answer:

<h2>They should make 8 small ones and 12 large ones.</h2><h2>$260 they will make in total.</h2>

Step-by-step explanation:

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5 0
3 years ago
If a manufacturer conducted a survey among randomly selected target market households and wanted to be 95​% confident that the d
katen-ka-za [31]

Answer:

We need a sample size of least 119

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

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The margin of error is:

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95% confidence level

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Sample size needed

At least n, in which n is found when M = 0.09

We don't know the proportion, so we use \pi = 0.5, which is when we would need the largest sample size.

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

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0.09\sqrt{n} = 1.96*0.5

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n = 118.6

Rounding up

We need a sample size of least 119

6 0
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