A simple 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 deviation of the dissolved oxygen content for the warehouse is known from longexperience to be aboutσ= 2 mg/L.1 (a) Find a 98% confidence interval for the unknown population mean dissolved oxygen content.(b) Interpret your interval.(c) What sample sizenis required to get a 98% confidence interval with error margin 0.5?

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:

[tex]\mu \pm z_{critical}\frac{\sigma}{\sqrt{n}}[/tex]

Putting the values, we get,

[tex]z_{critical}\text{ at}~\alpha_{0.02} = 2.33[/tex]

[tex]9.14 \pm 2.33(\frac{2}{\sqrt{44}} ) = 9.14 \pm 0.7025 = (8.4375,9.8425)[/tex]

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

[tex]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[/tex]

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