Question

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?

Answer 1

a) (8.4375,9.8425)

b) 87

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)(\sigma)/(√(n))`

Putting the values, we get,

`z_(critical)\text{ at}~\alpha_(0.02) = 2.33`

`9.14 \pm 2.33((2)/(√(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(\sigma)/(√(n))\\\\0.5 = 2.33(2)/(√(n))\\\\√(n) = (2.33* 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.