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The level of dissolved oxygen (DO) in a stream or river is animportant indicator of the water’s ability to support aquatic lifeand happens to be normally dis...

Question

The level of dissolved oxygen (DO) in a stream or river is animportant indicator of the water’s ability to support aquatic lifeand happens to be normally distributed. A researcher measures theDO level at 15 randomly chosen locations along a stream andcalculates the mean to be 4.7713 milligrams per liter (mg/l) andstandard deviation to be 0.9396 mg/l. A dissolved oxygenlevel below 5 mg/l puts aquatic life at risk. Can we conclude thataquatic life in this stream is at risk? Carry out a test att

The level of dissolved oxygen (DO) in a stream or river is an important indicator of the water’s ability to support aquatic life and happens to be normally distributed. A researcher measures the DO level at 15 randomly chosen locations along a stream and calculates the mean to be 4.7713 milligrams per liter (mg/l) and standard deviation to be 0.9396 mg/l. A dissolved oxygen level below 5 mg/l puts aquatic life at risk. Can we conclude that aquatic life in this stream is at risk? Carry out a test at the αα= 0.1 significance level to help you answer this question. (Links to an external site.) GeoGebra Probability Calculator (Links to an external site.) Step 1: State the claim and its opposite. Identify which is the null hypothesis and which is the alternative hypothesis. Step 2: Determine which hypothesis test you will use and check conditions. Step 3: If the conditions are met, perform the calculations, and conduct the test. Include the test statistic and the P-value. Step 4: State the result and interpret your result in the context of the problem. Please show work!



Answers

The quantity of dissolved oxygen (DO) in natural waters is an essential parameter for monitoring survival of most aquatic life. DO is affected by temperature and the amount of organic waste. An earlier method for determining DO involved a two-step process:
1. The water sample is treated with $\mathrm{KI}$,
$$\mathrm{O}_{2}(a q)+4 \mathrm{KI}(a q)+2 \mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow 2 \mathrm{I}_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{K}_{2} \mathrm{SO}_{4}(a q)$$
2. The $\mathrm{I}_{2}$ is titrated with sodium thiosulfate,
$$\mathrm{I}_{2}(a q)+2 \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}(a q) \longrightarrow \mathrm{Na}_{2} \mathrm{S}_{4} \mathrm{O}_{6}(a q)+2 \mathrm{NaI}(a q)$$
A 50.0 -mL water sample is treated with KI, and then 15.75 $\mathrm{mL}$ of 0.0105 $\mathrm{M} \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}$ is required to reach the end point.
(a) Which substance is oxidized in step 1, and which is reduced in step 2?
(b) What mass $(g)$ of $\mathrm{O}_{2}$ is dissolved in the water sample?

All right. So one condition for the distribution is that it needs to be normal in order to use the chi squared procedure. Um to the left or right, here is a graph of after plotting these, it looks like there's a couple of outliers in here, and that indicates to us that the distribution is not normal enough to use a chi square procedure, so we can't use we can't use it, therefore, it's not reason.

This question covers drawing normal distribution plots. So in order to draw a normal distribution plot, you must first know how much entry CSR there's 22 entries and we find the corresponding table three for 22 entries which begins with negative 1.91 from here. We can construct a table with these values sorted in order. So I would and put these in a calculator or some application that can sort numbers from highest, lowest. So the lowest point 17 and the highest I believe is 7.6. So you would sort this and find the corresponding um table three values. So the lowest value of table three corresponds to the lowest value from this dataset and so is the highest corresponds to the highest. And in between they also have the same pattern. Yeah, so once we have that we can figure out, we can plot it in an X. Y axis. Okay. Yeah. Once we plotted in an X Y axes we can determine two overall pattern of where the majority of the data is. And once we determine the overall pattern, it is pretty clear what data points are. Outliers. Yeah. So in this case there are two outliers right here and there are outliers because they don't follow the general trend, Right? So the outliers are 6.7 and 7.6 um Mimos per square meter per day of effusive oxygen uptake. Um And after this we can determine whether it is normally distributed or not. Yeah. So nose we normal since the majority of the data um follows this trend. If we just eliminated the outliers, then we know that it's mostly normal, so it's normal they distributed.

Hi everyone. So in this question, They're given the quantity of dissolved option. Do a national water is an essential parameter for monitoring survey of most acidic, like uh dissolved oxygen is affected by depression in the amount of organic question. Early method for determining the selection involved. The two step process first, the water sample is fitted with the K or two Plus four guy plus try sisters for accused toys I took plus traces to mosquitoes. Support Second, the idea started to the Syrian to yourself. It So I do like us plush toys. I need to s two or three years and it S 406 plush toys and a 50 ml of water uh sample is treated with the care and then 15.75 ml of europe and 010 for you, smaller and a two H 203 is required for uh to reach the in point resurgence is absolutely in the state one and research distances, reducing the steps to what mass production is dissolved in the water sample. So the role of flooring the city is Step one oxidizing substance. Mhm. Step one oxidize substance. So all too Plus four K. i. Loss. Yeah. H. two. So 4 gives so yes, I do less. No, he's H door lows can do a cell phone. Okay, so in step one K. Is oxidized. It's okay I it's oxidized. Okay, Its oxidation spread changes from -1-0. That is there is increasing oxidation number then step two. That's step do I. 20 Blatz boys. And they do as to or three gives a NATO as sport oh six plus toys. And may I. So in this case I do use reached I only is list from 0 to -1 then be yes. Yeah, Mosab or to dissolve in water. Moss of oh two. Be solved in water. So first, yeah. First we is equal to 15.75 ML. That is equal to 0.0 one for you. Seven for you leader. Yeah. Then yeah. Meese. According to 0.0 105 Moller. So the video or two In this whole water sample, W two in the salt water sample. They're difficult to. And into end that is equal to 0.00 1323 g. You get uh Thanks to the

We want to implement a left tail hypothesis test that is We want to test whether or not the population mean is less than the known means of new equals eight for this problem, we are given a sample with n equals 37 trials sampling the X. R equals 7.2 and sample standard deviation as equals 1.9. Gentlemen, this test, we want to use significance level alpha at 1%. To start off with, let's state the significance and hypotheses are significance. Is alpha equals 10.1 Our hypotheses are no new equals eight alternative H. M. You less than a What distribution will use. And let's compute associated testes cystic. Since we don't know the population sigma, We're gonna have to use a student's T distribution which is safe to use because we have angered at 30 satisfied given this we computer T stat given by this formula. Archie stat equates the negative 2.56 Next let's compute the p interval and sketch it out on the T distribution. Since we have degree of freedom and minus one equals 36. Will round that down to 35 in order to use the one to the one tail tea table. That is because the one lt table doesn't normally contain intervals of five for high degrees of freedom. So we round down to make sure as not to overestimate the p value. Okay, we find that our P value from the one tail tea table is between 10.5 point 01 That means that we can graph it as follows. We have Archie statin negative 2.56 on the T axis here and the p value is the very small area, the left of this cheese stat changing here in yellow. What can we conclude from this? Since r. P is less than or equal to our alpha value, we have statistically significant findings, Therefore, we can reject the null hypothesis, H not, and we can interpret this to me, we have sufficient evidence suggests MU is less than eight for this population.


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