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Lilit [14]
3 years ago
11

According to Gallup, 58% of US Adults choose to pursue higher education solely to get a good job. Consider taking samples of 900

adults in the United States and calculating the sample proportion who pursue higher education solely to get a good job.
Assuming all conditions are met, fill in the blanks for the following about the sampling distribution for the sample proportion.
The sampling distribution for the sample proportion follows the [ Select ] ["Population Model", "Sample Model", "Random Model", "Normal Model"] . The mean of the sampling distribution is [ Select ] ["0.064", "0.0165", "900", "0.58"] . The standard deviation of the sampling distribution is [ Select ] ["0.58", "0.064", "900", "0.0165"]
Mathematics
1 answer:
katovenus [111]3 years ago
4 0

Answer:

(a) Normal model

(b)\ Mean = 0.58

(c)\ \sigma = 0.0165

Step-by-step explanation:

Given

p = 58\%

n = 900

Solving (a): The distribution type

The sample follows a normal model

Solving (b): The mean

This is calculated as:

Mean = p

So, we have:

Mean = 58\%

Express as decimal

Mean = 0.58

Solving (c): The standard deviation

This is calculated as:

\sigma = \sqrt{\frac{p(1 - p)}{n}}

So, we have:

\sigma = \sqrt{\frac{58\%(1 - 58\%)}{900}}

Express as decimals

\sigma = \sqrt{\frac{0.58(1 - 0.58)}{900}}

\sigma = \sqrt{\frac{0.58 * 0.42}{900}}

\sigma = \sqrt{\frac{0.2436}{900}}

\sigma = \sqrt{0.00027066666}

\sigma = 0.0165

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hichkok12 [17]

Answer:

A. 95% confidence interval of gamble amount is (18.78277, 37.70227)

B. The 95% confidence interval of gamble amount is (42.23237, 100.3835)

C. 95% confidence interval of sqrt(gamble) is (3.180676, 4.918371)

D. The predicted bet value for a woman with status = 20, income = 1, verbal = 10, which shows a negative result and does not fit with the data, so it is inferred that model (c) does not fit with this information

Step-by-step explanation:

to)

We will see a code with which it can be predicted that an average man with income and verbal score maintains an appropriate 95% CI.

attach (teengamb)

model = lm (bet ~ sex + status + income + verbal)

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using the following formula you can obtain the confidence interval for the bet amount of 95%

predict (model, new data, range = "confidence")

lwr upr setting

28.24252 18.78277 37.70227

as a result, the confidence interval of 95% of the bet amount is (18.78277, 37.70227)

b)

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lwr upr setting

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we can deduce that a man with the maximum state, income and verbal punctuation is going to bet 71.30794

The 95% confidence interval of the bet amount is (42.23237, 100.3835)

it is observed that the confidence interval is wider for a man in maximum state than for an average man, it is an expected data because the bet value will be higher than the person with maximum state that the average what you carried s that simultaneously The, the standard error and the width of the confidence interval is wider for maximum data values.

(C)

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we replace:

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(d)

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lwr upr setting

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The predicted bet value for a woman with status = 20, income = 1, verbal = 10, which shows a negative result and does not fit with the data, so it is inferred that model (c) does not fit with this information

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