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Doss [256]
3 years ago
7

An automotive manufacturer wants to know the proportion of new car buyers who prefer foreign cars over domestic. Step 2 of 2 : S

uppose a sample of 1418 new car buyers is drawn. Of those sampled, 354 preferred foreign over domestic cars. Using the data, construct the 99% confidence interval for the population proportion of new car buyers who prefer foreign cars over domestic cars. Round your answers to three decimal places.
Mathematics
1 answer:
son4ous [18]3 years ago
7 0

Answer: 99% confidence interval would be (0.13,0.26).

Step-by-step explanation:

Since we have given that

Sample size n = 1418

x = 354

So, \hat{p}=\dfrac{x}{n}=\dfrac{354}{1418}=0.249

At 99% level of significance, z = 2.58

So, interval would be

\hat{p}\pm z\sqrt{\dfrac{p(1-p)}{n}}}\\=0.249\pm 2.58\times \sqrt\dfrac{0.249\times (1-0.249)}{1418}}}\\\\=0.249\pm 0.01148\\\\=(0.249-0.01148,0.249+0.01148)\\\\=(0.13,0.26)

Hence, 99% confidence interval would be (0.13,0.26).

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An article suggested that yield strength (ksi) for A36 grade steel is normally distributed with μ = 44 and σ = 5.0. (a) What is
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Answer:

Step-by-step explanation:

Using normal distribution,

z = (x - μ)/σ

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σ = standard deviation= 5.0

a) The probability that yield strength is at most 40=

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z = (40-44)/5= -0.8

Looking at the normal distribution table,

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b) P(x greater than 62) = 1 - P(x lesser than or equal to 62)

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