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The Effect of Carbohydrates on Proteolysis of Clycophorin (Integrates with Chapters $5,6,$ and $9 .$ ) Consider the sequence of glycophorin (see Figure 9.10 , and i...

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The Effect of Carbohydrates on Proteolysis of Clycophorin (Integrates with Chapters $5,6,$ and $9 .$ ) Consider the sequence of glycophorin (see Figure 9.10 , and imagine subjecting glycophorin, and also a sample of glycophorin treated to remove all sugars, to treatment with trypsin and chymotrypsin. Would the presence of sugars in the native glycophorin make any difference to the results?

The Effect of Carbohydrates on Proteolysis of Clycophorin (Integrates with Chapters $5,6,$ and $9 .$ ) Consider the sequence of glycophorin (see Figure 9.10 , and imagine subjecting glycophorin, and also a sample of glycophorin treated to remove all sugars, to treatment with trypsin and chymotrypsin. Would the presence of sugars in the native glycophorin make any difference to the results?



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For Exercises 5 through $20,$ assume that the variables are normally or approximately normally distributed. Use the traditional method of hypothesis testing unless otherwise specified. Carbohydrates in Fast Foods The number of carbohydrates found in a random sample of fast-food entrees is listed. Is there sufficient evidence to conclude that the variance differs from $100 ?$ Use the 0.05 level of significance. $$ \begin{array}{lllll}{53} & {46} & {39} & {39} & {30} \\ {47} & {38} & {73} & {43} & {41}\end{array} $$

In this problem. We're trying to determine if there's a statistically significant difference in the number of grams of carbohydrate in chocolate Candies versus non chocolate Candies, right? So are null. Hypothesis is that there's no difference is that the average number of carbs in chocolate Candies is the same as in non chocolate Candies. We're gonna test this against the alternative hypothesis that they're different without any specifications as to which direction we're expecting them to be different, one greater than the other. To make this test, we're gonna take two samples samples off chocolate Candies and samples of non chocolate Candies. We have our chocolate. Our sample of chocolates has 13 elements, and we cook a sample of 11 non chocolate Candies. The average number of grams of carbohydrate in the chocolate Candies was 29.7, and the average in the non chocolate Candies 34.4. We don't know the underlying population signals, and so we have to calculate a state or deviation from those samples as well. For the chocolate Candies, the the standard deviation was 6.5 grams, and for the non chocolate Candies, the standard deviation was 11.2 scrape up. Now the way we've structured are null and alternative hypotheses. This is going to be a two tailed test to t a two sided test, that is to say, and our test statistics for difference of means is, of course, x bar chocolate minus expert non chocolate. So the sample difference minus the difference in the expected values of the populations, standardized by dividing by a standard error which is based on the sample standard deviations. Since our no hypothesis is that the muse or the same then are expected value of the difference is zero, and this term drops out. So we have just this portion left. So this is our test statistic now because we are using the sample standard deviation and because we're able to use these smaller sample sizes because we're assuming that the underlying populations are normally distributed. This statistic can be assumed to be distributed according to the student T distribution, which, of course, means that we have to determine degrees of freedom. Degrees of freedom is simply the smaller of the two sample sizes minus one. In that case, in this case, it's going to be 11 minus one or 10 so we have 10 degrees of freedom, and the problem specifies that we air to use a 10% or Alfa equals 100.1 significance level. So if I have a two sided test with a T distribution with 10 degrees of freedom and critical values defined by an Alfa of 0.1 critical values for this test, we're going to be plus or minus 1.8125 and visualizing that the most artistic this is the region such that if they are test statistic, if rt statistic falls within this range, we will not reject the null hypothesis. RT statistic has to fall out in one of these two tails in order to reject and all high prosthesis. As it turns out, if you calculated the T statistic equals negative 1.22 here's our zero point, and that's approximately there. There's our statistic, and it is within the region where we do not reject the null hypothesis. Which is to say we have no reason to assume on a statistically significant basis that non chocolate Candies differ from chocolate Candies in terms of the number of grams of card carbohydrate per one ounce serving

What's up, stat cats. In this video, we are given an example and asked to perform a one way nova. And if we have a significant result, we are then asked to perform either a chef or to keep post talk test. So let's go ahead and look at the raw data we were given. And this is the fiber content between different types of foods and we're told, are awful. Level is 0.5 So let's go ahead and do our one way Unova Unova Single Factor on Excel. Our input range is all of our data, and we do have labels in the first row and are out level is 0.5 very important. Okay, so this is the output for our know the test. And what we want to look at is are critical value in our test value. So if our test statistic is greater than our critical value, then we have a significant unova results. So that would mean at least one of thes means is different from each other between groups. So is 1.99 larger than three point for seven? No, it's not so we don't have a significant results. So consequently, we don't have a reason to perform any sort of post talk test. None of our means are significantly different from each other. So we failed to reject are no hypothesis. So there is not enough evidence to conclude that at least one of our means is different from each other. And because we don't have a significant result, there is no need to perform a post hoc test. All right, you guys, that's it for this video. I hope you learn. Go out. I'll see next time.

Hello. Everyone said they were going to talk about riel, wife example of family dynamics, which is digestion. So if we think about eating a complex sugar or carbohydrate, let's say it looks like this and her body breaks it down into simple sugars, which, with an individual like this, we could then say that there is an increase. An entropy do two more particles being created.

So what happened in this experiment was that in the presence of the Yuria, the wrong sulfide bonds or die sulfide bombs, the wrong dice vilified bonds. We're prepared. Okay, so, um, the wrong ones repaired because there are. There's over 100 different ways of pairing the eight Sistine molecules. So over 100 technically 100 and five total. Um, that Onley one combination. One combo is in dramatically active, hasn't it? Is an enzyme that can do something. One combo is active in dramatically. So if there are all these wrong combination, do you have 100 over 100 pairs of red over 100 pairs? That can happen. And you have this, um, Yuria this ammonia that's causing the wrong pairs to be formed. It makes sense that you're gonna only have 1%. Um, that's that's in dramatically active because the rest of them have basically just been scrambled together. And don't do anything. Um, so, yeah, that's how you would get the 1%


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