n=800 x=632 Example: The National Association of Music Merchants surveyed 800 parents of children 5-18, and 632 of them said that music education has a positive effect on academic performance. Assuming these parents to be a random sample, we would like to construct a 95% confidence interval for the proportion of parents who believe that music education has a positive effect on academic performance. One example by hand for understanding - you will use your calculator when finding confidence intervals (see next slide) Check the assumptions and conditions. ✓ We have a simple random sample. The sample size of 800 is less than 10% of the population of all parents. ✓ It's safe to assume that the individuals within the sample are independent. ✓ The sample contains 632 successes and 168 failures 800-632

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Where did they get the 0.21 from? Please explain 

n=800
x=632
Example: The National Association of Music Merchants surveyed 800 parents of children 5-18, and 632 of
them said that music education has a positive effect on academic performance. Assuming these parents to
be a random sample, we would like to construct a 95% confidence interval for the proportion of parents
who believe that music education has a positive effect on academic performance.
One example by hand for understanding - you will use your calculator when finding confidence intervals (see next slide)
Check the assumptions and conditions.
✓ We have a simple random sample.
✓ The sample size of 800 is less than 10% of the population of all parents.
✓ It's safe to assume that the individuals within the sample are independent.
✓ The sample contains 632 successes and 168 failures
800-632
95% CI = p + z ² √ B²
632
800
C-level = 0.95 => z* = 1.96
p =
= 0.79
= 0.79 ± 1.96 (0.79) (0.21)
(0.79)
800
= 0.79 ± 1.96 (0.0144)
+
= 0.79 ± 0.0282 → (0.7618, 0.8182)
ME
Transcribed Image Text:n=800 x=632 Example: The National Association of Music Merchants surveyed 800 parents of children 5-18, and 632 of them said that music education has a positive effect on academic performance. Assuming these parents to be a random sample, we would like to construct a 95% confidence interval for the proportion of parents who believe that music education has a positive effect on academic performance. One example by hand for understanding - you will use your calculator when finding confidence intervals (see next slide) Check the assumptions and conditions. ✓ We have a simple random sample. ✓ The sample size of 800 is less than 10% of the population of all parents. ✓ It's safe to assume that the individuals within the sample are independent. ✓ The sample contains 632 successes and 168 failures 800-632 95% CI = p + z ² √ B² 632 800 C-level = 0.95 => z* = 1.96 p = = 0.79 = 0.79 ± 1.96 (0.79) (0.21) (0.79) 800 = 0.79 ± 1.96 (0.0144) + = 0.79 ± 0.0282 → (0.7618, 0.8182) ME
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