3. A simple random sample was taken of 44 water bottles from a bottling plant’s warehouse. The dissolved oxygen 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, 7.34 Here the sample mean is 9.14 mg/L. The population standard deviation of the dissolved oxygen content for the warehouse is known from long experience to be about σ = 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 size n is required to get a 98% confidence interval with error margin 0.5?

a) The 98% confidence interval for the unknown population mean dissolved oxygen content is between 8.44 mg/L and 9.84 mg/L.

b) This interval means that we are 98% sure that the true population mean dissolved oxygen content is between 8.44 mg/L and 9.84 mg/L.

c) A sample size of 87 is required to get a 98% confidence interval with error margin 0.5

Step-by-step explanation:

We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of .

So it is z with a pvalue of , so

Now, find M as such

In which is the standard deviation of the population and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 9.14 - 0.7 = 8.44 mg/L.

The upper end of the interval is the sample mean added to M. So it is 9.14 + 0.7 = 9.84 mg/L.

The 98% confidence interval for the unknown population mean dissolved oxygen content is between 8.44 mg/L and 9.84 mg/L.

b) This interval means that we are 98% sure that the true population mean dissolved oxygen content is between 8.44 mg/L and 9.84 mg/L.

(c) What sample size n is required to get a 98% confidence interval with error margin 0.5?

This is n for which M is 0.5. So

Rouding up

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