Question

their nearest polling station. then state whether the confidence interval you construct contradicts the reporter’s claim. (if necessary, consult a .)
(a) click on \take sample\ to see the results for your random sample.

enter the values of the sample size, the point estimate of the mean, the sample standard deviation, and the critical value you need for
your 95% confidence interval. (choose the correct critical value from the table of critical values provided.) when you are done, select
\compute\.

( t_{0.005} = 2.807 )
( t_{0.010} = 2.500 )
( t_{0.025} = 2.069 )
( t_{0.050} = 1.714 )
( t_{0.100} = 1.319 )

(b) based on your sample, graph the 95% confidence interval for the population mean of all the distances the voters in rising falls county
live from their nearest polling station.

• enter the values for the lower and upper limits on the graph to show your confidence interval.
• for the point (•), enter the claim 4.64 from the article.

(c) does the 95% confidence interval you constructed contradict the claim made in the article?
choose the best answer from the choices below.
○ no, the confidence interval does not contradict the claim. the mean of 4.64 km from the article is
inside the 95% confidence interval.
○ no, the confidence interval does not contradict the claim. the mean of 4.64 km from the article is
outside the 95% confidence interval.
○ yes, the confidence interval contradicts the claim. the mean of 4.64 km from the article is inside the
95% confidence interval.
○ yes, the confidence interval contradicts the claim. the mean of 4.64 km from the article is outside
the 95% confidence interval.

Explanation:

Step1: Identify sample values

Sample size $n=24$, sample mean $\bar{x}=4.864$, sample standard deviation $s=2.421$, degrees of freedom $df=24-1=23$. Critical value for 95% confidence (two-tailed, $\alpha/2=0.025$) is $t_{0.025}=2.069$.

Step2: Calculate standard error

Standard error $SE=\frac{s}{\sqrt{n}}=\frac{2.421}{\sqrt{24}}\approx\frac{2.421}{4.899}\approx0.494$.

Step3: Compute margin of error

Margin of error $ME=t_{0.025}\times SE=2.069\times0.494\approx1.022$.

Step4: Find confidence interval

Lower limit: $\bar{x}-ME=4.864-1.022\approx3.842$
Upper limit: $\bar{x}+ME=4.864+1.022\approx5.886$

Step5: Check claim consistency

The reporter's claim is 4.64, which is inside (3.842, 5.886).

Answer:

(a) Sample size: 24, Point estimate: 4.864, Sample standard deviation: 2.421, Critical value: 2.069
(b) Lower limit ≈3.84, Upper limit≈5.89, Point at 4.64
(c) No, the confidence interval does not contradict the claim. The mean of 4.64 km from the article is inside the 95% confidence interval.