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Sampletablet) Result from Q#8 4.976 1.663 5.334tablet) Result from Q#8 23.67 13.91 14.08Calculate the mean, standard dcviation, rclative standard deviation and true...

Question

Sampletablet) Result from Q#8 4.976 1.663 5.334tablet) Result from Q#8 23.67 13.91 14.08Calculate the mean, standard dcviation, rclative standard deviation and true value range at a 95% confidence interval (Has%c )of mass (mg) of pyridoxine and thiamine per average Swiss Natural stress vitamin tablet Match the statistics to the correct valucs_

Sample tablet) Result from Q#8 4.976 1.663 5.334 tablet) Result from Q#8 23.67 13.91 14.08 Calculate the mean, standard dcviation, rclative standard deviation and true value range at a 95% confidence interval (Has%c )of mass (mg) of pyridoxine and thiamine per average Swiss Natural stress vitamin tablet Match the statistics to the correct valucs_



Answers

Anemia For the study of Jordanian children in Exercise 64, the sample mean hemoglobin level was
11.3 mg/dl and the sample standard deviation was 1.6 mg/dl.
(a) Calculate the test statistic.
(b) Find the P-value using Table B. Then obtain a more precise P-value from your calculator.

Uh huh. We want to test that new hypothesis. H Not new equals 1000. Against the alternative. You? Greater than 1000. For sample with size and equals 50. Expire equals about 1081 as equals 426. Where the data comes from a normally distributed population. And we conduct the test that alpha equals 4260.5 conference. This question is testing the knowledge of how to conduct a hypothesis. Test the population menu with a standard deviation of the population segment is unknown. We proceed in three steps to dissolve. First day we come to the test that this is Tina equals x minus mu not overactive library and not as 1000. So plugging in the above data gives 1.344 Next we compute the critical value for our Alpha. We need to use a tea table with degree of freedom 49 to find the T. Value for which the area to the right of the students the distribution of 1000 doing so, we find T. C. Equals 1.677 since Tina is less than T. C. And this is the right tool tests. We conclude that Tina is not in the critical region, which means we c. We conclude we fail to reject because Tina is not in the critical region.

We are very familiar with the methods. By now we want to construct the confidence level of 99 point Sorry off 99% and we have 20 different readings. So in order to control the confidence interval, what I've done is I've again simply input the values over here. These you can see we have 20 values. We are example mean a study 2.5 and a standard deviation is 20.3, you know, confidence level of 99. I select 99 this is the confidence interval that I get 20.8 to 44.2 20.8 20 pointed 20.8 to 44 point to This is my 99% confidence in trouble. Alright. Could also have done this by using the same old formula here and would have become 20 degree of freedom would be 19. Substitute these values Alfa. I would have become 0.50 point 00 fight and put the values in this formula and we come to the same result. This one now it is asked. Does this say? What is the question? There's the conference interval. Give us a good information about the population of all cancers off the same pretty brands that I consumed. No, because we have taken only one can off every brand in this, and only one can cannot be a representative off all the cans off that particular brand. So the answer is no. And we don't even care if this is normally distributed, because this is simply not going to give us a good idea off the, you know, off the mean off the mean off the mean of what we are. Yeah, the mean off caffeine for 12 rounds off a drink. So we don't even care if this is normally distributed.

Let us read this question. A fast food restaurant estimates that the mean sodium content in one of its breakfast sandwiches is no more than 9 20. It is no more than 9 20 which means mu is less than or equal to 9 20. So this is going to be my null hypothesis. And what will be my alternative hypothesis? It will be that new is greater than nine 20 a random sample of 44 breakfast sandwiches. So I'm simply going to write the Formula 44. So in the denominator, this will be route off 44. I'm using the Zed Statistic, so random sample of 44 breakfast sandwiches has a mean off 9 25. So I mean, there is Expert is 9 25. So this is going to be 9 25 minus 9 20 minus muse, which is the hypothesized mean upon what is the standard deviation. It is 18 mg if there is 18. So if I use a calculator for this, this would be five divided by 18 multiplied by route off 44. This is 1.84 to my sense statistic is 1.842 I see that this is I am checking for greater than so this is a right tail test, A right tail test. So what is my Alfa my Alfa 0.1? So what is going to be my critical value? My critical value offset statistic for 0.1 is 1.282 My critical value is 1.282 If this is my critical value of 1.28 to my that statistic off, 1.8 falls here, which means I will reject my null hypothesis. I reject h not. This means what? Let us go to the question. A fast food restaurant claims that the means story in content in one off its breakfast sandwiches is no more than 9 20. So we say that we have enough statistical evidence to say that the claim off the fast food restaurant is wrong. We have enough statistical evidence to say that the mean sodium content in that particular breakfast sandwich, or the mean off on what we can say is the main sodium content in those breakfast sandwiches is more than 920 and this would be my answer

In this exercise, we're going to be considering a random sample that has gotten from a population of non standard deviation, which equals 22.1. So the population standard deviation is given As 22.1 And we're supposed to get a 95% confidence interval for the population mean? For two given to given cases in fast case a were given that The sample size and is 121 and the sample mean is 82 0.4. And it also given that the sample standard deviation, He is 21 0.9. So we're going to use this data to determine the 95% confidence interval for the mean. So we proceed as follows, we have to use um the formula The sample mean plus or minus the critical value that corresponds to the 95% confidence interval And that is one 0.96 zero times the sample Senate division divided by the square root of And so let's go on and substitute the values. So we have X. Bar which is 82.4 plus or minus 1.960 Multiplied by 22.1 divided by the square root of 121. So when you work that out you get 82.4 plus or minus 3.94. When rounded off to the nearest to two decimal places in party. The problem we're working with a sample size N of 81 a sample mean X. Bar of 82 .4 as well. And the sample standard deviation of 21.9. So we still substitute the formula that we had there in red Which is 82.4 class or minus. Yeah. Um 1.960 multiplied by 22 0.1 divided by the square root off The sample size of 81. And when you walk it up you get 82.4 Plus or -4.81. Now, in both cases, we can conclude that their population mean is going to be within Deranged, outlined by the 95% confidence interval for the population mean, and notice that we did not use the sample standard deviation 21.9 in both cases. And this is because the population standard division Is known and it has given us 22.1. Mhm.


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