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 the results below.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 and the sample standard deviation is 1.78 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.

(a) Find a 98% confidence interval for the unknown population mean dissolved oxygen content.

(b) Interpret your interval.

Answer 1

Answer:

a)$9.14-2.33\frac{2}{\sqrt{44}}=8.44$    

$9.14+2.33\frac{2}{\sqrt{44}}=9.84$    

b) The 98% confidence interval would be given by (8.44;9.84)    

For this case we can conclude that we have 98% of confidence that the true mean for the oxygen content is between 8.44 and 9.84 mg/L

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=9.14$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

$\sigma=2$ represent the population standard deviation

n=44 represent the sample size  

Part a

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$   (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=18-1=17$

Since the Confidence is 0.98 or 98%, the value of $\alpha=0.01$ and $\alpha/2 =0.01$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.01,0,1)".And we see that $z_{\alpha/2}=2.33$

Now we have everything in order to replace into formula (1):

$9.14-2.33\frac{2}{\sqrt{44}}=8.44$    

$9.14+2.33\frac{2}{\sqrt{44}}=9.84$    

Part b

The 98% confidence interval would be given by (8.44;9.84)    

For this case we can conclude that we have 98% of confidence that the true mean for the oxygen content is between 8.44 and 9.84 mg/L