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Answer :
Jamie needs a sample size of 1181 (rounded up from 1180.8 as we can't have a partial patient) to estimate the proportion with the desired accuracy.
3) Jamie needs to figure out how many households she needs to sample.
She wants her estimate to be within $25 of the population mean with a 95% confidence level.
If she assumes that the population standard deviation is $288, she can use the formula for the sample size when estimating a mean:
[tex]n = (Z-score \times \sigma / E)^2[/tex]
Where:
Z-score refers to the value from the standard normal distribution corresponding to the desired confidence level (for a 95% confidence level, the Z-score is approximately 1.96).
σ is the standard deviation of the population, $288 in this case.
E is the desired margin of error, $25 in this case.
Substituting these values into the formula:
[tex]n = (1.96 \times 288 / 25)^2[/tex]
n = 134.5
Since we cannot have a fraction of a household, Jamie needs to sample 135 households to achieve her desired accuracy and confidence level.
4a) Mrs. Cole found that 32 out of 120 students sampled plan on getting a four-year or higher college degree in science or math fields.
We first calculate the sample proportion (p-hat): p-hat = 32/120 = 0.267
Then we calculate the standard error (SE):
[tex]SE = \sqrt{[ p-hat \times ( 1 - p-hat) / n ]}[/tex]
[tex]=\sqrt{[ 0.267 \times(1 - 0.267) / 120 ] = 0.041}[/tex]
Now we find the Z-score for a 90% confidence level, which is 1.645.
The margin of error (ME) is Z-score [tex]\times[/tex] SE: ME = 1.645 [tex]\times[/tex] 0.041 = 0.067
So the margin of error at a 90% confidence level is approximately 0.067 or 6.7%.
4b) A 90% confidence interval for the true proportion is calculated as follows:
Lower bound = p-hat - ME = 0.267 - 0.067 = 0.20
Upper bound = p-hat + ME = 0.267 + 0.067 = 0.334
So the 90% confidence interval is (0.20, 0.334).
4c) The lower confidence interval is greater than 0.22, hence we can say that the percentage of students planning to pursue the four-year or higher degree in science or math fields is greater than 22%.
5) The orthodontist wants to estimate the proportion of his patients interested in an alternative to traditional braces with a 98% confidence level and wants the estimate to be within 3.4% of the true proportion.
If there's no preliminary estimate, we need to use the formula for the sample size:
[tex]n = (Z-score^2 \times p \times (1 - p))/E^2[/tex]
Let p = 0.5 (If no preliminary estimate is available, we use 0.5).
The Z-score for a 98% confidence level is approximately 2.33.
Substituting these values into the formula:
[tex]n = (2.33^2 \times 0.5 \times(1 - 0.5))/(0.034^2)[/tex]
n = 1180.8
For similar question on sample size.
https://brainly.com/question/28583871
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