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Joan's Nursery specializes In custom-dcsigned landscaping for residential areas. The estimated Iabor cast associated wlth- paricular landscaping proposal based...

Question

Joan's Nursery specializes In custom-dcsigned landscaping for residential areas. The estimated Iabor cast associated wlth- paricular landscaping proposal based on the number plantIngs tees, shrubs; and so on t0 0e used foretha project . For cost-estimating pumpascs managers use two hours Iabor lme for the planting ofa medium-sized trze_ Actual tlmes from sample ot 10 plantings during the past month follow (tlmes hours)With 0.05 level of significance, test see whether the mean tree planting

Joan's Nursery specializes In custom-dcsigned landscaping for residential areas. The estimated Iabor cast associated wlth- paricular landscaping proposal based on the number plantIngs tees, shrubs; and so on t0 0e used foretha project . For cost-estimating pumpascs managers use two hours Iabor lme for the planting ofa medium-sized trze_ Actual tlmes from sample ot 10 plantings during the past month follow (tlmes hours) With 0.05 level of significance, test see whether the mean tree planting time differs from hvo hours State the null and altemative hypotheses. Ho: p 2 Ho: 0>2 Ha: # $ 2 "2 2 Compute the sampl Mcam Compute the sample standard deviation ruuno Your Jnjwer three decimnal places.) What the test statistic? (Round vour answer Lhree decimal pleces:) What the p-value? (Round your answer t0 four decimal places,) p-value What = your conclusion? Do not reject Ho: We can conclude that the mean Frce-d anting time differs from two hours There reason change from the tWo hours (or cost estimating aurposes Do not reject Ha' Wc cannot conclude that thc mcun trce-planting timc dilfers from two hours. There Is no Feason change from the tWo hours for cost estimating purposes: Reject Ho: We can conclude that the mean tree-planting tIme difters from two hours: There [.0 reason t0 change from the two hours for cost estimating purposes; Reject Ho: We cannot conclude that the mean tree-planting time differs from two hours_ There no fedson to change from the two hours for cost estmating purposes-



Answers

Joan's Nursery specializes in custom-designed landscaping for residential areas. The estimated labor cost associated with a particular landscaping proposal is based on the number of plantings of trees, shrubs, and so on to be used for the project. For costestimating purposes, managers use two hours of labor time for the planting of a medium-sized tree. Actual times from a sample of 10 plantings during the past month follow (times in hours).
$$\begin{array}{lllllllll}{1.7} & {1.5} & {2.6} & {2.2} & {2.4} & {2.3} & {2.6} & {3.0} & {1.4} & {2.3}\end{array}$$
With a .05 level of significance, test to see whether the mean tree-planting time differs from two hours.
a. State the null and altemative hypotheses.
b. Compute the sample mean.
c. Compute the sample standard deviation.
d. What is the $p$ -value?
e. What is your conclusion?

The following is a nova test based on the mean salaries for different metropolitan areas. So the alternative or the null hypothesis is that all the means are the same. So there are six metropolitan areas, I think it goes Chicago, Dallas Miami, Denver san Diego and Seattle. Uh So the null hypothesis is that all the means are the same. And then the alternative is that at least one of them is different. The second step is to find the critical value and you can do that using either software or a table, But they're essentially three things you need. The first thing is your alpha value, your significance level and that's usually given to you the problem and that's .05. Then you need the degrees of freedom for the numerator and the degrees of freedom for the denominator. And the way you find that Is the degrees of freedom for the numerator is the number of categories -1. So there were six cities that we looked at our metropolitan areas, so 6 -1° of freedom would be five for the numerator. And then for the denominators, the total number of data values minus the number of categories. So there were 36 data values minus the six metropolitan areas. So 30 is your degrees of freedom for the denominator. So that should be enough to use a table. But I use a calculator and I wrote a program in here called inverse. F. I'm not going to show you how to how to write the program. You can youtube it if you wish. Um But this is what I do. So um I put in my area which is my alpha value, my degrees of freedom is five and then my degrees of freedom for the denominator is 30 and That gives me my critical value. About 2.534 2534 is my critical value. I call f. star. So 2.534. Okay so anything greater than 2.534. We reject the annual hypothesis that all the means are the same And anything less than 2.534. We failed to reject meaning the h not is true. Okay so the second step is to find the F statistic and there's a formula but it's a bit of a mess. I always use software you know technology is a great thing. So if you go to stat and you can type in your data values. So these are the mean salaries um So again L1 I think was Chicago and then this is the mean salary for Dallas Miami Denver San Diego and Seattle. So there are six categories. And if you go to stat tests and then we're gonna go to the Unova test and then you just type in your columns separated by commas remember there were six columns, six data columns that we used and we need to make sure that all of them are in there and last one and then also you know make sure you separate those by commons, otherwise it's going to read it wrong. So then um that gives us everything we need. So the F. Is the F statistic, that's the third step. So we're looking at this it's about 2.281 as our F. Value. So two point 281 is our f statistic Which is actually barely in the non rejection region 2.281. So that means we fail to reject. Okay and also we can verify that with this p value here. So the p values 0.7 which is a pretty small p value, but it's still in this case greater than the alpha value. So the alpha value remembers point oh five, so it's barely greater than the alpha value. And whenever it's greater than the alpha value, uh we failed to reject, I should probably put H not there, so we failed to reject H not whenever the P values greater than the alpha. Okay. So then the last step is to summarize everything with actual words. So what does this all mean? It just means that there is not sufficient evidence, there is not sufficient. I guess you could say statistical evidence to suggest that the mean salaries from the different metropolitan areas are different. Okay. And that's the five step process for an Innova one way and over test

We want to conduct a pair differences test at the alpha equals 5% level testing the claim that population mean X bar A is greater than population X barbie. We have data a be given here, we assume amounts to mr distribution as you can see on the right. I've already calculated the mean difference D bar 6.125 The sample size and eight and the sample standard deviation of differences SD and 8.7 We complete the five steps us to blow to solve this problem first, let's evaluate the requirements to use a student's T distribution of the hypotheses because of the distribution shape it is appropriate to use a student distribution your degree of freedom and minus 27. No hypothesis mute equals zero. Alternative media is greater than zero and alpha equals 00.5 for confidence nexus, complete the test at and the P value our test that is T equals D. Bar over SDR. Again this gives 2.14 U. T. Table. This gives us a P value between 0.5 point 025 That means we can include that P is less than equal to alpha. So we reject the null hypothesis H not which means that we have evidence and you D. Is greater than zero.

We have muse equal to 1300 as a known mean of the population. We want to calculate the sample mean exploring the sample standard deviation s for the randomly sampled data that follows. So to do so, we simply have to remember the definition of expire to ask for a sample. That's why, for example, in some of the data to buy buy em in this case, 12 68 S is some of the deviations about the moon squared, all divided by n minus one. In this case. 37.29 Next, what we want to do is implement a two tailed test. That is we want to test whether or not this sample suggests the actual meaning this population differs either higher or lower. Either direction from the no mean 1300. We're going to do so with an alcohol level of 0.1 and we're gonna know at this point that X is approximately normally distributed. So to implement this test, we have to follow the following procedures, one by one 1st. What is the significance of hypotheses? Alpha equals 0.1 The null hypothesis H not is musical 1300. And the alternative hypothesis H is that it is not equal 1300. Next distribution We're going to use compute the associated testes, cystic. We're going to use a student's t distribution because the population standard deviation sigma is not known. We only have a sample standard deviation S It's appropriate to use the distribution because X is approximately normally distributed, meaning it's both symmetric and round shape necessary to use students T. Six reasons to distribution. Let's calculate the T statistic. It's defined by this formula, expire minus mu divided by s over root end, which in this case equates the negative 2.714 next let's compute the p interval and sketch the results. So, since we have a degree of freedom of n minus one equals nine, no, from a two tailed T table, which we can find in google and the stats textbook, we find that the T statistic corresponds to a p value range between 20.0 to 1.5 We can think of this also as the area underneath the T distribution outside of negative 2.714 and 2.714 The T statistic as is graft on the right. Finally, given this p interval range, we can make a conclusion about this test. Since P is greater than alpha, we have statistically insignificant bindings and we cannot reject h shot. We fail to a general hypothesis. Therefore, we interpret this to mean that we lack evidence suggesting that the population mean differs from its no mean of 1300.

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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