Question

Six measurements were made of the magnesium ion concentration (in parts per million, or ppm) in a city's municipal water supply, with the following results. It is reasonable to assume that the population is approximately normal.
70 201 199 202 173 153

Based on a 95% confidence interval for the mean magnesium ion concentration, is it reasonable to believe that the mean magnesium ion concentration may be greater than 208? (Hint: you should first calculate the 95% confidence interval for the mean magnesium ion concentration.)

A. No

B. The likelihood cannot be determined.

C. Yes

Answer 1

Answer:

Yes

Step-by-step explanation:

1) 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".

Dataset given: 70, 201 ,199, 202, 173 ,153

We can calculate the sample mean and deviation with the following formulas:

$\bar X = \frac{\sum_{i=1}^n x_i}{n}=166.33$

$s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}=51.075$

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

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n=6 represent the sample size  

2) Confidence interval

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

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

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

$df=n-1=6-1=5$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,5)".And we see that $t_{\alpha/2}=2.57$

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

$166.33-2.57\frac{51.075}{\sqrt{6}}=112.742$    

$166.33+2.57\frac{51.075}{\sqrt{6}}=219.918$

So on this case the 95% confidence interval would be given by (112.742;219.918)    

Is it reasonable to believe that the mean magnesium ion concentration may be greater than 208?

On this case since the confidence interval contains the value 208 and the values above 208 until 219.918 makes sense the statement.

YES

We can conduct an hypothesis test

We need to conduct a hypothesis in order to check if the mean is less than 208, the system of hypothesis would be:  

Null hypothesis:$\mu \geq 208$  

Alternative hypothesis:$\mu < 208$  

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

$t=\frac{166.33-208}{\frac{51.075}{\sqrt{6}}}=-1.998$    

P-value

The first step is calculate the degrees of freedom, on this case:  

$df=n-1=6-1=5$  

Since is a one side left tailed test the p value would be:  

$p_v =P(t_{(5)}$  

Conclusion  

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL reject the null hypothesis, so we fail to reject the hypothesis that the mean is higher or equal than 208 at 5% of significance.