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Sav [38]
3 years ago
10

In a large population of adults, the mean IQ is 112 with a standard deviation of 20. Suppose 200 adults are randomly selected fo

r a market research campaign. The distribution of the sample mean IQ is (a)exactly Normal, mean 112, standard deviation 20. (b)approximately Normal, mean 112, standard deviation 0.1. (c)approximately Normal, mean 112, standard deviation 1.414. (d)approximately Normal, mean 112, standard deviation 20. (e)exactly Normal, mean 112, standard deviation 1.414.
Mathematics
1 answer:
Snezhnost [94]3 years ago
4 0

Answer:

(c)approximately Normal, mean 112, standard deviation 1.414.

Step-by-step explanation:

To solve this problem, we have to understand the Central Limit Theorem

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

In this problem, we have that:

\mu = 112, \sigma = 20, n = 200

Using the Central Limit Theorem

The distribution of the sample mean IQ is approximately Normal.

With mean 112

With standard deviation s = \frac{20}{\sqrt{200}} = 1.414

So the correct answer is:

(c)approximately Normal, mean 112, standard deviation 1.414.

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To obtain information on the corrosion-resistance properties of a certain type of steel conduit, 45 specimens are buried in soil
Debora [2.8K]

Answer:

statistic value t= 5.43

p-value: < 0.00001.

Step-by-step explanation:

Hello!

To obtain information over the corrosion-resistance properties of a certain type of steel conduit a random sample of 45 specimens was taken and buried for two years.

The study variable is:

X: Max. penetration of a steel conduit.

The data of the sample

n= 45

sample mean X[bar]= 53.4

sample standard deviation S= 4.2

The conduits are manufactured to have a true average penetration of at most 50 mills, symbolically: μ ≤ 50

The hypothesis is:

H₀: μ ≤ 50

H₁: μ > 50

α: 0.05

To choose the corresponding statistic to use to study the population mean, the variable must have a normal distribution. There is no available information to check this, so I'll just assume that the variable has a normal distribution and, since the population variance is unknown and the sample is small, the statistic to use is a Student t.

Under the null hypothesis, the critical region and the p-value are one-tailed.

Critical value:

t_{n-1; 1-\alpha } = t_{44; 0.95} = 1.68

Rejection rule:

Reject the null hypothesis when t ≥ 1.68

t= \frac{53.4 - 50}{\frac{4.2}{\sqrt{45} } }

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The calculated value is greater than the critical value, thedecision is to rject the null hypothesis.

p-value:

P(t ≥ 5.43) = 1 - P(t < 5.43) = < 0.00001.

The p-value is less than α so the decision is to reject the null hypothesis.

Since the null hypothesis was rejected, then the population average of the penetration of the conduits specimens is greater than 50 mils. It is not recommendable to use these conduits.

I hope it helps!

4 0
3 years ago
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