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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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6. A bank auditor claims that credit card balances are normally distributed, with a mean of $3570
sergejj [24]

Answer:

Probability that their mean credit card  balance is less than $2500 is 0.0073.

Step-by-step explanation:

We are given that a bank auditor claims that credit card balances are normally distributed, with a mean of $3570  and a standard deviation of $980.

You randomly select 5 credit card holders.

Let<em> </em>\bar X<em> = </em><u><em>sample mean credit card balance</em></u>

The z score probability distribution for sample mean is given by;

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where, \mu = population mean credit card balance = $3570

            \sigma = standard deviation = $980

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Now, the probability that their mean credit card  balance is less than $2500 is given by = P(\bar X<em> </em>< $2500)

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The above probability is calculated by looking at the value of x = 2.44 in the z table which has an area of 0.9927.

Therefore, probability that their mean credit card  balance is less than $2500 is 0.0073.

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Answer:

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Hi!

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Have a nice day!

7 0
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Read 2 more answers
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