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The principal of a large school claims that the average distance from a student’s home to the school is 3.5 km - Leaving Cert Mathematics - Question 2 - 2015
Question 2
The principal of a large school claims that the average distance from a student’s home to the school is 3.5 km. In order to test this claim, a sample of 60 students ... show full transcript
Worked Solution & Example Answer:The principal of a large school claims that the average distance from a student’s home to the school is 3.5 km - Leaving Cert Mathematics - Question 2 - 2015
Step 1
Test the principal’s claim using a 5% level of significance. Clearly state your null hypothesis, your alternative hypothesis and your conclusion.
Answer
To test the principal's claim:
Null Hypothesis (H₀): The mean distance from the students’ homes to the school is 3.5 km (H₀: µ = 3.5)
Alternative Hypothesis (H₁): The mean distance from the students’ homes to the school is not 3.5 km (H₁: µ ≠ 3.5)
Next, we calculate the z-score:
egin{align*}
ext{Sample Mean} (ar{x}) & = 3.7 ext{ km}
ext{Population Mean} ( ext{µ}) & = 3.5 ext{ km}
ext{Standard Deviation} (σ) & = 0.5 ext{ km}
ext{Sample Size} (n) & = 60
ext{Standard Error (SE)} & = rac{ ext{σ}}{ ext{√n}} = rac{0.5}{ ext{√60}} = 0.0645
ext{Z} & = rac{ar{x} - ext{µ}}{ ext{SE}} = rac{3.7 - 3.5}{0.0645} = 3.1
ext{Rejection Regions: } & z ext{ is less than } -1.96 ext{ or greater than } 1.96
\end{align*}
Since 3.1 lies in the rejection region, we reject the null hypothesis at the 5% level of significance.
Step 2
Find the 95% confidence interval for the proportion of students from that school who live within 2.5 km of the school.
Answer
Given that 20% of the 60 students live within 2.5 km:
Sample Proportion (ar{p}): 0.2 Sample Size (n): 60
The confidence interval is calculated as follows:
egin{align*}
CI & = ar{p} ext{ ± } Z imes ext{SE}
ext{where } ext{SE} & = ext{√} rac{ar{p}(1 - ar{p})}{n} = ext{√} rac{0.2(0.8)}{60} ext{ = } 0.0646
Z & = 1.96 ext{ for 95% confidence}
\end{align*}
Thus the confidence interval is:
egin{align*} (0.2 - 1.96 imes 0.0646, 0.2 + 1.96 imes 0.0646) & = (0.0988, 0.3012) \end{align*}
95% Confidence Interval: (0.0988, 0.3012)
Step 3
Based on your answer to part (b) is it possible to conclude, at the 5% level of significance, that the proportion of students living within 2.5 km of the school has changed since that time? Explain your answer.
Answer
Null Hypothesis (H₀): The proportion of students living within 2.5 km is 26% (0.26)
Alternative Hypothesis (H₁): The proportion of students living within 2.5 km is different from 26% (0.26)
We calculate the z-score:
egin{align*}
ext{Sample Proportion} (ar{p}) & = 0.2
ext{SE} & = ext{√} rac{0.26(1 - 0.26)}{60} = 0.061
\text{Z} & = rac{0.2 - 0.26}{0.061} = -0.98
\end{align*}
At the 5% level of significance, the rejection region is between -1.96 and 1.96. Since -0.98 does not fall in this range, we cannot reject the null hypothesis. Thus, we cannot conclude that the proportion of students living within 2.5 km has changed.
Step 4
What is the minimum sample size necessary for the statistician to carry out this analysis?
Answer
To estimate the proportion with 95% confidence and a margin of error (E) of 10% (0.1):
The formula for sample size is:
egin{align*}
n & = rac{Z^2 imes p(1 - p)}{E^2}
ext{Using } p = 0.5 ext{ (most conservative estimate)}:
n & = rac{(1.96)^2 imes 0.5(0.5)}{(0.1)^2} = rac{(3.8416) imes 0.25}{0.01}
& = 96.04
\end{align*}
Thus the minimum sample size necessary is 97 (rounded up).