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15)According to the 444 Daily Fuel Gauge Report; the state of California average price for regular unleaded gasoline on January 2nd; 2018 was S3.11.A week later; ra...

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15)According to the 444 Daily Fuel Gauge Report; the state of California average price for regular unleaded gasoline on January 2nd; 2018 was S3.11.A week later; random sample of 12 gas stations across Orange County yielded the following prices for - regular unleaded gasoline:s3.093.19 3.09 3.193.04 2.95 335 3.15333Use normal probability plot to assess whether the sample data above could have come from population that is normally distributed_ The data appear to be approximately normally distribu

15) According to the 444 Daily Fuel Gauge Report; the state of California average price for regular unleaded gasoline on January 2nd; 2018 was S3.11.A week later; random sample of 12 gas stations across Orange County yielded the following prices for - regular unleaded gasoline:s 3.09 3.19 3.09 3.19 3.04 2.95 335 3.15 333 Use normal probability plot to assess whether the sample data above could have come from population that is normally distributed_ The data appear to be approximately normally distributed € B) The data does not appear to be normally distributed 16) A simple random sample of size =40 is obtained from normal population with L = 50 and 0 =4. What is the sampling distribution of T 24 normally distributed with mean of 50 and standard deviation of 0.632 B) normally distributed with mean of 50 and standard deviation of 44 approximately normally distributed with mean of 50 and standard deviation of 0.6324 8 approximately normally distributed with mean of 50 and standard deviation of 44 shape unknown with mean of 50 and standard deviation of 44 17) The amount of money collected by snack bar at a large university has been recorded daily for the past five ~ years_ Records indicate that the mean daily amount collected is $3900 and the standard deviation is S600. Thex distribution skewed to the right due to several high volume days (including football game days) Suppose- that 100 days were randomly selected from the five ears and the average amount collected from those days ~ was recorded_ Which of the 'following describes the sampling distribution of the sample mean? - normally distributed with mean of S3900 and standard deviation of S60 B) normally distributed with mean of S3900 and standard deviation of S600e C) approximately normally distributed with mean of S3900 and standard deviation of $604 D) skewed to the right with mean of $3900 and standard deviation of S600~ E) skewed to the right with mean of $3900 and standard deviation of S60- 18) The reasoning for your answer in Problem #17 is: = the law of large numbers ~ the central limit theorem- 1 the 68-95-99_ rule the fact that probability is the long-run proportion of times an event occurs - the sampling distributione



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Please provide the following information. (a) What is the level of significance? State the null and alternate hypotheses. (b) Find the value of the chi-square statistic for the sample. Are all the expected frequencies greater than $5 ?$ What sampling distribution will you use? What are the degrees of freedom? (c) Find or estimate the $P$ -value of the sample test statistic. (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis that the population fits the specified distribution of categories? (e) Interpret your conclusion in the context of the application. Meteorology: Normal Distribution The following problem is based on information from the National Oceanic and Atmospheric Administration (NOAA) Environmental Data Service. Let $x$ be a random variable that represents the average daily temperature (in degrees Fahrenheit) in July in the town of Kit Carson, Colorado. The $x$ distribution has a mean $\mu$ of approximately $75^{\circ} \mathrm{F}$ and standard deviation $\sigma$ of approximately $8^{\circ} \mathrm{F}$. A 20 -year study $(620$ July days) gave the entries in the rightmost column of the following table. (i) Remember that $\mu=75$ and $\sigma=8 .$ Examine Figure $6-5$ in Chapter $6 .$ Write a bricf explanation for Columns I, II, and III in the context of this problem. (ii) Use a $1 \%$ level of significance to test the claim that the average daily July temperature follows a normal distribution with $\mu=75$ and $\sigma=8$

We want to conduct the pair differences. Test that alpha equals 5% confidence. Testing the claim that population mean expert A. Is greater than population index. Barbie. Were given pair data samples A and B. Here and we assume amount, shapes and distribution on the right. Have already concluded or rather computed the differences mundi bar sample size and and the standard deviation differences SD as 4.5, 10 and 4.12 respectively. With this info we can proceed to do five texas below to compute this test first. That the value of the requirements using student's T. Distribution as well as hypotheses. So the requirements on that because the distribution specified degree freedom and mine as well as nine are null and alternative hypotheses are nude equals zero moody greater than zero and we have alpha equals 00.5 is our confidence. Next. It's gonna be the test at the P value. The test that is T equals D. Bar over sdo route and equals 3.45 And the tea table. This gives P between 0.5 point 0005 Thus we can include since P is equal to alpha, that we reject the null hypothesis H not which means that we have evidence that moody is greater than zero.

We want to conduct a pair differences test at alpha equals 5% significance testing. The claim that the population means X. A and X. B are not equal. The data is given below. With mount shapes and at your distribution on the right. I've already computed D. Bar 2.25 The sample size N equals eight and the standard deviation S. D equals 7.78 We proceeded to five steps listed below to solve first. We'll check the requirements and evaluate hypotheses. So because the distribution shape requires have been met to use the student's T distribution, the degree of freedom is at minus 27. We have no hypothesis. H. And R equals zero alternative mu D. Does not equal zero Alpha equals 00.5 significance. Archer statistic is T equals D. Bar over SD over route and 4.8181 from the tea table R. P value is 0.5 is less than ph 7.25 So we have enough information to conclude our test. Now, since P is greater than Apple, we fail to reject the null hypothesis, which means that we lack evidence that beauty does not equal zero.

We have data sets A and B on the right. We want to utilize these data sets to test the claim that there is no difference in the underlying distributions for A and B it out the equals 0.1 significance. This question is testing an understanding of non parametric tests, particularly how to take perform the ranks on tests we proceed through steps A through G to solve. So first we see the outside of hypotheses. These are alpha equals 0.1 hypotheses. H. And distributions are saying hk distribution are different and be we can be the test at the sampling distribution which is normal and check the requirements which are met. So their ranks are for A and B. As follows That we have and one equals 11 and two equals 12. You are sigma are 1 30 to 16.2 from this, we have our equals and summer range from a 1 73 so Z equals ar minus. Moreover signal articles 2.53 thus received complete the P value. As for normal distribution two PZ greater than 0.114 Thus, we conclude, indeed, that we fail to reject nation on since P is greater than alpha, which means we lack evidence for AJ.

What we want to conduct A p D. T. A pair differences tests at the alpha equals 1% confidence level. Testing the claim that the population means A bar next pr are not equal. We have the data for A and B. Given below assuming amount shapes, not your distribution on the right. I've already calculated D. Bar noted that an equal seven and calculated SD as 70.47 So we proceeded to do the five steps listed below to solve this first. We evaluate the requirements and hypotheses. So the requirements to use the students distribution have been met because the distribution shape we have degree of freedom and minus one equals six. Are null hypothesis is mute equals zero. Or alternative is beauty does not equal zero. And we're testing at alpha equals 00.1 confidence Next will compute the test statistic and P value. So the statistic is T equals D. Bar divided by SD over. Uh huh. 2.083 from a tea table. This puts p between .1.05. So we can conclude that P is greater than alpha, which means we fail to reject the null hypothesis, which means that we lack evidence that beauty does not equal to about.


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