The masses, $$\(m\)$$ kilograms, of flour in a random sample of 90 sacks of flour are summarised as follows. $$\( n=90 \quad \Sigma m=4509 \quad \Sigma m^{2}=225950 \)$$ Calculate a $$\(98 \%\)$$ confidence interval for the population mean. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................
Exam No：9709_s21_qp_63 Year：2021 Question No：4(b)
Answer:
\(' 50.1 ' \pm z \sqrt{\frac{\frac{491}{890}}{90}}\)
\(z=2.326\)
\(49.9\) to \(50.3(3 \mathrm{sf})\)
\(z=2.326\)
\(49.9\) to \(50.3(3 \mathrm{sf})\)
Knowledge points:
6.4.7 determine and interpret a confidence interval for a population mean in cases where the population is normally distributed with known variance or where a large sample is used
Solution:
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