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djverab [1.8K]
3 years ago
6

One way to check on how representative a survey is of the population from which it was drawn is to compare various characteristi

cs of the sample with the population characteristics. A typical variable used for this purpose is age. The 2010 GSS of the American adult population found a mean age 49.28 years and a standard deviation of 17.21 for its sample of 4,857 adults. Assume that we know from Census data that the mean age of all American adults is 37.2 years.
Required:
a. State the research and the null hypothesis setting for a two-tailed test.
b. Calculate the t statistics and test the null hypothesis setting alpha at .01. What did you find?
c. What is your decision about the null hypothesis? What does this tell us about how representative the sample is of the American adult population?
Mathematics
1 answer:
weqwewe [10]3 years ago
7 0

Answer:

a) See step by Step explanation

b) z(s) = 48.88

c) We reject H₀. The sample is not representative of American Adult Population

Step-by-step explanation:

From sample

sample mean .    x = 49.28

sample standard deviationn   s = 17.21

sample size  n₁  = 4857

Population mean according to Census data

μ  = 37.2

a) Test Hypothesis

Null Hypothesis .                  H₀ .                    x =  μ  = 37.2

Alternative Hypothesis        Hₐ .                    x ≠ μ

b) We have sample size (4857) we can use normal distribution

z (c) for    α = 0.01   α/2 . = 0.005  is from z-table . z(c) = 2.575

To calculate  z(s) =  ( x  - μ ) / s /√n

z(s) =  12.08 * √4857 / 17.21

z(s) = 12.08* 69.64 / 17.21

z(s) = 48.88

z(s) > z(c)

We should reject H₀. The sample is not representative of American Adult population

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Answer:

The mean is 9.65 ohms and the standard deviation is 0.2742 ohms.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

10% of all resistors having a resistance exceeding 10.634 ohms

This means that when X = 10.634, Z has a pvalue of 1-0.1 = 0.9. So when X = 10.634, Z = 1.28.

Z = \frac{X - \mu}{\sigma}

1.28 = \frac{10.634 - \mu}{\sigma}

10.634 - \mu = 1.28\sigma

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5% having a resistance smaller than 9.7565 ohms.

This means that when X = 9.7565, Z has a pvalue of 0.05. So when X = 9.7565, Z = -1.96.

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1.96\sigma + 1.28\sigma = 10.645 - 9.7565

3.24\sigma = 0.8885

\sigma = \frac{0.8885}{3.24}

\sigma = 0.2742

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\mu = 10.634 - 1.28\sigma = 10 - 1.28*0.2742 = 9.65

The mean is 9.65 ohms and the standard deviation is 0.2742 ohms.

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