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A production line operates with a mean filling weight of 16 ounces per container. Overfilling or underfiling presents a serious problem and when detected requires the operator to shut down the production line to readjust the filling mechanism. From past data, a population standard deviation $\sigma=.8$ ounces is assumed. A quality control inspector selects a sample of 30 items every hour and at that time makes the decision of whether to shut down the line for readjustment. The level of significance is $\alpha=.05 .$a. State the hypothesis test for this quality control application.b. If a sample mean of $\overline{x}=16.32$ ounces were found, what is the $p$ -value? What action would you recommend?c. If a sample mean of $\overline{x}=15.82$ ounces were found, what is the $p$ -value? What action would you recommend?d. Use the critical value approach. What is the rejection rule for the preceding hypothesis testing procedure? Repeat parts (b) and (c). Do you reach the same conclusion?

a. $H_{0} : \mu=16, H_{a} : \mu \neq 16$b. $P=0.0286$ , Readjustment is neededc. $P=0.2186,$ No readjustment is neededd. The same conclusions are reached.

Intro Stats / AP Statistics

Chapter 9

Hypothesis Tests

Confidence Intervals

Hypothesis Testing with One Sample

Hypothesis Testing with Two Samples

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and this problem. We're trying to see whether there is a problem in filling the the filling this container, and if there's over or under feeling, there is a problem. So because there's over or under feeling, we know that our alternative hypothesis is that mu is not equal to 16 ounces because if it was just under filling, then we could save me was less than 16 ounces if it was just over feeling we could save mu is greater than 16 ounces. But because we're saying overfilling and under feeling presents a problem, our music, not our alternative hypothesis, is going to be that meat was not equal to 16 ounces. That means or no hypothesis is going to be amused equal to 16 ounces. Also given in the problem is that we have a population standard deviation of zero point eight, and we have a sample size of 30 items. So this is the answer to part A. These are our hypotheses in Part B were given that we have an ex bar of 16.32 so we have to find a P value in order to find a P value we need to compute a test statistic because we have a ah, population standard deviation, That's what. This this Ingrid here represents this pop. We have a population standard deviation. We can compute ese test statistic just equal to x bar minus the population mean over. Ah, population. Standard deviation divided by the square root of end is equal to 16.32 minus 16 divided by 0.8 but by the square root of 30. And once we compute those values, we get a Z score of 2.19 And now we plot this one a curve. This normal curve is equal zeros in the middle. Izzy equals 2.19 Occurs the right side on. We're interested in finding this area over here, Um, everything to the right. And under the curve from Z equals 2.19 This value represents the area represents probability that Z is greater than or equal to two. 2.19 And we can find this value from, um by taking this area over here, I'm subtracting it from one. So this is equal to one, minus the probability that sea is less than or less than 2.19 which is equal to one minus zero point 9857 and we get a P value or a probability of 0.0 one for three. However, this is not our p value, because our mu is not equal to something. So we have a to tail to tail test or two tailed hypothesis tests s. So that means that our P value is going to be two times probability that c two times the probability that Z is greater than or equal to 2.19 which is equal to 0.286 So when we compare our p value of 0.286 to our Alfa of 0.5 um, 3.286 you're a 0.286 is less than 0.5 Therefore, we can reject the no hypothesis in part C. We do the same thing with a different sample. Mean our sample, mean is now 15.82 15.8, 15.82 and we in order to find a P value, we need to compute a test statistic because again, we're using a population standard deviation, we need to computer ese test statistic Z is equal to x bar, which is 15.82 minus our population mean, which is 15 over our population. Standard deviation of 0.8 divided by the square root of our sample size, which is 30. This value is equal to 1.232 And when we plot this onto a distribution begin, C equals zero over here as equals, 1.232 curves to the right of that and we're interested in finding the area of the curve under the area under the curve and to the right of Z equals 1.23 To this, this area represents a probability that Z is greater than or equal to 1.232 and we can find this value by subtracting the area to the left of Z equals 1.232 from one. So one minus the probability that he is less than 1.232 which is equal to one point one minus point eight nine 17 I believe one minus 10.8917 And when you get that theater value of 0.1093 But this is not our p value, because again, we have a two tailed test statistics. So this is just half of our P value. You have to multiply this by two. So our P value is equal to zero point 2186 And now we compare our P of 0.2186 to our Alfa of 0.5 and we get that we fail to reject No, because a 0286 is greater than their important 05 And lastly, when we are discussing, um, our rejection rule. So since we have a two tailed Z test, um, we're going to have to split our Alfa into two parts. We have, um, over too on the right and Alfa over two on the left. So we have to find a Z associated with Alfa over too. Um, and in this situation are, Alfa is equal to 0.5 Alfa over to is equal to 0.0 to 5. And at a Alfa of 0.0 to 5 with the probability Significance elbow 0.25 we have a Z score of one 1.96 So this is our Z score. So in order to reject the the no to reject the no, um, we must have a Z value. We must have a Z score greater than 1.96 Or is he less than 1.96 in order for us to reject the knoll at a 0.5 significance level? And in this situation, because be calculated, who were here? A t statistic of 2.192 point 19 in part, be 2.19 is greater than 1.96 So we can reject the note. And in part, see, um oh, this must be is he is less than negative. 1.96 use less than negative 1.96 Um, over here in port. See, we have a Z score of 1.232 So 1.232 is less than 1.96 So we failed to reject the no

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