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Andre45
1 month ago
9

According to a Field Poll conducted February 8 – 17, 2005, 79% of California adults (actual results are 400 out of 506 surveyed)

feel that "education and our schools" is one of the top issues facing California. We wish to construct a 90% confidence interval for the true proportion of California adults who feel that education and the schools is one of the top issues facing California.
Find a 90% confidence interval for the population proportion. (Round your answers to three decimal places.)

_______to_______
Mathematics
1 answer:
Inessa [12.5K]1 month ago
7 0

Answer: (0.760, 0.820)

Step-by-step explanation:

Denote p as the proportion of adults in California who believe that "education and our schools" are among the foremost concerns in the state.

Given: 79% (actual results indicate 400 out of 506 surveyed) of adults in California consider "education and our schools" one of the main issues currently facing the state.

In other words, sample size: n= 506

Sample proportion: \hat{p}=0.79

Critical value for a 90% confidence interval (Using z-value table):

z=1.645

Now, the confidence interval for the population proportion at 90% will be:

\hat{p}\pm z\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}

In other words, 0.79\pm (1.645)\sqrt{\dfrac{0.79(1-0.79)}{506}}

In summary, 0.79\pm (1.645)(0.0181)

\approx0.79\pm 0.030=(0.79-0.030,\ 0.79+0.030)\\\\=(0.760,\ 0.820)

Thus, the 90% confidence interval for the population proportion= (0.760, 0.820)

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