Mathematics: Asimple random sample was taken of 44 water bottles from a bottling plant’s warehouse. the dissolvedoxygen content (in mg/l) was measured for each bottle, with these results: 11.53, 8.35, 11.66, 11.54, 9.83, 5.92, 7.14, 8.41, 8.99, 13.81, 10.53, 7.4, 6.7, 8.42, 8.4, 8.18, 9.5, 7.22, 9.87,6.52, 8.55, 9.75, 9.27, 10.61, 8.89, 10.01, 11.17, 7.62, 6.43, 9.09, 8.53, 7.91, 8.13, 7.7, 10.45, 11.3, 10.98, 8.14,11.48, 8.44, 12.52, 10.12, 8.09, 7.34here the sample mean is 9.14 mg/l.the population standard

I estamted 137Step-by-step explanation:...

Asimple random sample was taken of 44 water bottles

Ответ:

jackson5637

07.01.2022

Mathematics

a) (8.4375,9.8425)

b) 87

Step-by-step explanation:

We are given the following data-set.

11.53, 8.35, 11.66, 11.54, 9.83, 5.92, 7.14, 8.41, 8.99, 13.81, 10.53, 7.4, 6.7, 8.42, 8.4, 8.18, 9.5, 7.22, 9.87,6.52, 8.55, 9.75, 9.27, 10.61, 8.89, 10.01, 11.17, 7.62, 6.43, 9.09, 8.53, 7.91, 8.13, 7.7, 10.45, 11.3, 10.98, 8.14,11.48, 8.44, 12.52, 10.12, 8.09, 7.34

n = 44

Sample Mean = 9.14 mg per Litre

Population Standard Deviation = 2 mg per Litre

a) Formula:

Confidence interval:

$\mu \pm z_{critical}\frac{\sigma}{\sqrt{n}}$

Putting the values, we get,

$z_{critical}\text{ at}~\alpha_{0.02} = 2.33$

$9.14 \pm 2.33(\frac{2}{\sqrt{44}} ) = 9.14 \pm 0.7025 = (8.4375,9.8425)$

b) Thus, it could be said that above interval gives about 98% confidence that it will contain the true value of mean of the dissolved oxygen content for the warehouse.

c) Margin of error

$z_{critical}\displaystyle\frac{\sigma}{\sqrt{n}}\\\\0.5 = 2.33\frac{2}{\sqrt{n}}\\\\\sqrt{n} = \frac{2.33\times 2}{0.5} = 9.32\\\\n = 86.86 \approx 87$

Thus, a sample size of 87 is required to get a 98% confidence interval with error margin 0.5.

0,0(0 оценок)

Ответ:

Ответ:

Ask your question to the AI advisor

More tips

Answers on questions: Mathematics

Ask an AI advisor a question