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alukav5142 [94]
2 years ago
13

A statistics lecturer poses the following question to her students as homework: 'Suppose I collected a sample and calculated the

sample proportion. If I construct a 90% confidence interval for the population proportion and a 95% confidence interval for the population proportion, which of these intervals will be wider?' Three students provide their answers as follows: Tim: 'The 90% confidence interval will be wider.' Trevor: 'The 95% confidence interval will be wider.' Tracy: 'There is not enough information to tell. Either interval could be wider.'
Mathematics
1 answer:
zhenek [66]2 years ago
3 0

Answer: The 95% confidence interval will be wider.

Step-by-step explanation:

Confidence interval for population proportion is written as

Sample proportion ± margin of error

margin of error = z score × √pq/n

The z score is determined by the confidence level. The z score for a confidence level of 95% is higher than the z score for a confidence level of 90%

This means that with all other things being equal, a 95% confidence level will give a higher margin of error compared to a 90% confidence level.

The higher the margin of error, the wider the confidence interval. Therefore,

The 95% confidence interval will be wider.

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An organization has members who possess IQs in the top 4% of the population. If IQs are normally distributed, with a mean of 100
murzikaleks [220]

Answer:

The minimum IQ score will be "126".

Step-by-step explanation:

The given values are:

Mean

\mu = 100

Standard deviation

\sigma=15

Now,

⇒ P(z>x)=4 \ percent \ i.e., 0.04

⇒ P(z>\frac{x- \mu}{\sigma} )=0.04

⇒ 1-P(z \leq \frac{x-100}{15})=0.04

⇒ \frac{x-100}{15}=z0.96

            =NORMDIS(z=0.96)

            =1.751

⇒ x=100+15\times 1.751

      =126.265 \ i.e., 126

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2 years ago
25 POINTS!!!!!!
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Your anwer is C)<span>f(x) = 467(5 to the one fourth power)4x</span>
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