# Practice Problems 9

### Problem 1: A Muslim president?

A survey of 1,527 American adults conducted in June 2015 stated that 60% would vote for a qualified Muslim presidential candidate. The survey goes on to say that “… the margin of sampling error is +/- 3 percentage points at the 95% confidence level.”

(a). What is the relevant sample statistic? Give appropriate notation and the value of the statistic.##
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*Answer:*\(\hat{p} = 0.60\)

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*Answer:*p = the proportion of all American adults who vote for a qualified Muslim candidate

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*Answer:*\(0.60 \pm .03\) gives an interval from 0.57 to 0.63 I am 95% confident that the proportion of American adults who would vote for a Muslim presidential candidate is between 57% and 63%.

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*Answer:*The proportion of all Americans who would vote for a Muslim presidential candidate is likely between 57% and 63%, so we could say that majority (>50%) would vote for a Muslim candidate.

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*Answer:*About 95% of all samples of 1527 American adults will give us a sample proportion who would vote for a Muslim presidential candidate that is within 3% of the population proportion who would vote for a Muslim presidential candidate.

### Problem 2: Biomass in Tropical Forests

Using a random sample of 4079 inventory plots, scientists found a sample average of 11,600 tons of carbon per square kilometer with a standard error of 1000 tons. Give a 95% confidence interval for the mean amount of carbon per square kilometer in tropical forests. Clearly interpret the meaning of this confidence interval.

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*Answer:*\(11,600 \pm 2(1000)\) gives an interval from 9,600 to 13,600. We are 95% sure that the mean amount of carbon per square kilometer in all tropical forests is between 9,600 and 13,600 tons.

### Problem 3: Change in gun ownership?

A 2016 study described in The Guardian found that a random sample of US adults in 1994 found a female rate of gun ownership of 9%. A similar random sample in 2015 found the rate of female gun ownership rose to 12%. Though not given in the article, let’s assume that the SE for the difference in these two sample proportions is 2%.

(a). Use correct notation to describe our parameter of interest: the difference in the proportion of female gun owners in 1994 and 2015.##
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*Answer:*\(\hat{p}_{1994} - \hat{p}_{2015}\)

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*Answer:*\(\hat{p}_{1994} - \hat{p}_{2015}\) = 0.09 - 0.12 = -0.03

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*Answer:*\((0.09 - 0.12) \pm (.02) = -0.03 \pm 0.04 = -0.07 \text{ to } 0.01, or -7\% \text{ to } 1\%\)

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*Answer:*We are 95% confident that the female gun owner rate in 1994 could be 7 percentage point lower to 1 percentage point higher than the rate in 2015. We can say that the observed increase from 1994 to 2015 of 3% is not “statistically significant” because it is within the margin of error (4%) for this study.

### Problem 4: Interpreting a Confidence Interval

Using a sample of 24 deliveries described in “Diary of a Pizza Girl” on the Slice website, we find a 95% confidence interval for the mean tip given for a pizza delivery to be $2.18 to $3.90. Which of the following is a correct interpretation of this interval? Indicate all that are correct interpretations.

(a). I am 95% sure that all pizza delivery tips will be between $2.18 and $3.90.##
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*Answer:*Incorrect. The interval is about the mean, not individual tips..

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*Answer:*I am 95% sure that the mean pizza delivery tip for this sample will be between $2.18 and $3.90.

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*Answer:*Correct!

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*Answer:*Incorrect. The confidence is in where the population mean is, not where the interval itself is.