W75A - Please answer in detail using EXCEL Function

T-test of 2 means (Similar to presentation in class; but in a different context)

              (Wednesday class)

              Researchers in Georgia take a sample of diameters from 30 old-growth pine trees growing on two separate and adjoining tracts of land, labeled north and south, to see if the trees are identical.   The data are shown in the Tables below and reproduced in the Excel answer workbook.

1. Using Excel’s t-test of two means, calculate the P-value of a two-tail taste to identify whether the mean tree diameter from the north tract and south tracts are identical or not at a 5% level of significance. (Round to four digits.) Make sure you state the null and alternative hypotheses, as well as a formal conclusion to your test. (The Excel answer workbook reminds you to do so!) Assume variances (and, therefore, standard deviations are not equal. (1 point)

(North)

27.8	14.5	39.1	3.2	58.8	55.5	25	5.4	19	30.6
15.1	3.6	28.4	15	2.2	14.2	44.2	25.7	11.2	46.8
36.9	54.1	10.2	2.5	13.8	43.5	13.8	39.7	6.4	4.8

              (South)

44.4	26.1	50.4	23.3	39.5	51	48.1	47.2	40.3	37.4
36.8	21.7	35.7	32	40.4	12.8	5.6	44.3	52.9	38
2.6	44.6	45.5	29.1	18.7	7	43.8	28.3	36.9	51.6

Problem 1:

Strategies for treating hypertensive patients by nonpharmacologic methods are compared by establishing three groups of hypertensive patients who receive the following types of nonpharmacologic therapy: Group 1:   Patients receive counseling for weight reduction Group 2:   Patients receive counseling for meditation Group 3:   Patients receive no counseling at all The reduction in diastolic blood pressure is noted in these patients after a 1-month period and are given in the table below. Group 1 Group 2 Group 3 4.2 4.5 1.2 6.4 2.3 −0.3 3.4 2.3 0.1 2.3

Work through Example 13 (which starts on p. 36 of the lab manual) and then do the following using the data from Problem #1 above. (a) Find the p-value for the hypothesis test in Problem #1(a) above. (b) Construct a normal probability plot of the residuals which includes the p-value for the normality test (similar to Figure 9, p. 41 of the lab manual). Enter the p-value into the answer box. (c) Find the p-value for the hypothesis test of equality of population variances (using Bartlett's test).

B. The proportion of customers who are completely satisfied in a recent satisfaction survey of 300 customers at XYC Inc. is found to be 0.26. Test the hypothesis that the population proportion of customers who are completely satisfied is greater than 0.22 using the critical value approach and a 0.05 level of significance. e Test the hypothesis that the population proportion of customers who are completely satisfied is less than 0.30 using the p-value approach and a 0.05 level of significance. b. Test the hypothesis that the population proportion of customers who are completely satisfied is different from 0.24 using the p-value approach and a 0.05 level of significance. C.

If you have a chance please answer as many as possible, thank you and I really appreciate your help experts!

Question 16 2 pts

In a hypothesis test, the claim is μ≤28 while the sample of 29 has a mean of 41 and a standard deviation of 5.9. In this hypothesis test, would a z test statistic be used or a t test statistic and why?

t test statistic would be used as the sample size is less than 30

t test statistic would be used as the standard deviation is less than 10

z test statistic would be used as the mean is less than than 30

z test statistic would be used as the sample size is greater than 30

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Question 17 2 pts

A university claims that the mean time professors are in their offices for students is at least 6.5 hours each week. A random sample of eight professors finds that the mean time in their offices is 6.2 hours each week. With a population standard deviation of 0.49 hours, can the university’s claim be supported at α=0.05?

No, since the test statistic is in the rejection region defined by the critical value, the null is rejected. The claim is the null, so is not supported

Yes, since the test statistic is not in the rejection region defined by the critical value, the null is not rejected. The claim is the null, so is supported

Yes, since the test statistic is in the rejection region defined by the critical value, the null is not rejected. The claim is the null, so is supported

No, since the test statistic is not in the rejection region defined by the critical value, the null is rejected. The claim is the null, so is not supported

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Question 18 2 pts

A credit reporting agency claims that the mean credit card debt in a town is greater than $3500. A random sample of the credit card debt of 20 residents in that town has a mean credit card debt of $3619 and a standard deviation of $391. At α=0.10, can the credit agency’s claim be supported?

Yes, since p-value of 0.09 is less than 0.55, reject the null. Claim is alternative, so is supported

No, since p-value of 0.09 is greater than 0.10, fail to reject the null. Claim is alternative, so is not supported

Yes, since p-value of 0.19 is greater than 0.10, fail to reject the null. Claim is null, so is supported

No, since p-value of 0.09 is greater than 0.10, reject the null. Claim is null, so is not supported

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Question 19 2 pts

A car company claims that its cars achieve an average gas mileage of at least 26 miles per gallon. A random sample of eight cars from this company have an average gas mileage of 25.6 miles per gallon and a standard deviation of 1 mile per gallon. At α=0.06, can the company’s claim be supported?

No, since the test statistic of -1.13 is close to the critical value of -1.24, the null is not rejected. The claim is the null, so is supported

Yes, since the test statistic of -1.13 is not in the rejection region defined by the critical value of -1.77, the null is not rejected. The claim is the null, so is supported

Yes, since the test statistic of -1.13 is not in the rejection region defined by the critical value of -1.55, the null is rejected. The claim is the null, so is supported

No, since the test statistic of -1.13 is in the rejection region defined by the critical value of -1.77, the null is rejected. The claim is the null, so is not supported

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Question 20 2 pts

A researcher wants to determine if extra homework problems help 8^th
grade students learn algebra. One 8^th grade class has extra homework problems and another 8^th grade class does not. After 2 weeks, the both classes take an algebra test and the results of the two groups are compared. To be a valid matched pair test, what should the researcher consider in creating the two groups?

That the group without extra homework problems receives different instruction

That the group with the extra homework problems has fewer after school activities

That each class has similar average IQs or abilities in mathematics

That each class of students has similar ages at the time of the testing

The manufacturer claims that your new car gets 31 mpg on the highway. You suspect that the mpg is a different number for your car. The 40 trips on the highway that you took averaged 28.7 mpg and the standard deviation for these 40 trips was 5.8 mpg. What can be concluded at the α α = 0.01 level of significance? a.For this study, we should use Select an answer z-test for a population proportion t-test for a population mean b.The null and alternative hypotheses would be: H0: H0: ? μ p ? ≠ = > < H1: H1: ? μ p ? > < = ≠ c.The test statistic ? t z = (please show your answer to 3 decimal places.) d.The p-value = (Please show your answer to 4 decimal places.) e.The p-value is ? ≤ > α α f.Based on this, we should Select an answer reject fail to reject accept the null hypothesis. g.Thus, the final conclusion is that ... The data suggest that the sample mean is not significantly different from 31 at α α = 0.01, so there is statistically insignificant evidence to conclude that the sample mean mpg for your car on the highway is different from 28.7. The data suggest that the population mean is not significantly different from 31 at α α = 0.01, so there is statistically insignificant evidence to conclude that the population mean mpg for your car on the highway is different from 31. The data suggest that the populaton mean is significantly different from 31 at α α = 0.01, so there is statistically significant evidence to conclude that the population mean mpg for your car on the highway is different from 31. h.Interpret the p-value in the context of the study. There is a 1.64116434% chance of a Type I error. If the population mean mpg for your car on the highway is 31 and if you take another 40 trips on the highway then there would be a 1.64116434% chance that the population mean would either be less than 28.7 or greater than 33.3. There is a 1.64116434% chance that the population mean mpg for your car on the highway is not equal to 31. If the population mean mpg for your car on the highway is 31 and if you take another 40 trips on the highway, then there would be a 1.64116434% chance that the sample mean for these 40 highway trips would either be less than 28.7 or greater than 33.3. i.Interpret the level of significance in the context of the study. There is a 1% chance that you own an electric powered car, so none of this matters to you anyway. If the population population mean mpg for your car on the highway is different from 31 and if you take another 40 trips on the highway, then there would be a 1% chance that we would end up falsely concluding that the population mean mpg for your car on the highway is equal to 31. There is a 1% chance that the population mean mpg for your car on the highway is different from 31. If the population mean mpg for your car on the highway is 31 and if you take another 40 trips on the highway, then there would be a 1% chance that we would end up falsely concluding that the population mean mpg for your car on the highway is different from 31.

1) The red blood cells counts of women are normally distributed with a mean of 4.577 and a standard deviation of 0.382.

Find the probability that a randomly selected woman has a red blood cell count that is lower than 4.2.

(Results are rounded to 4 decimal places)

a) 0.1618

b)  0.8382

c)  0.3382

d) 0.7979

2) A KRC research poll asked respondents if they felt vulnerable to identity theft. Of 1002 people polled, 531 said "yes".

Find a 95% confidence interval for the proportion of people who felt vulnerable to identity theft.

(Results are rounded to 4 decimal places)

a)  (0.5034, 0.5564)

b)  (0.4990, 0.5608)

c)   (0.4769, 0.5829)

d)   (0.5041, 0.5559)

Use repetitions to solve it. Don't simplify answer

(1) 20 different comic books will be distributed to five kids. How many ways are there to distribute the comic books if they are divided evenly so that 4 go to each kid?

(2) A family has four daughters. Their home has three bedrooms for the girls. Two of the bedrooms are only big enough for one girl. The other bedroom will have two girls. How many ways are there to assign the girls to bedrooms?

(3) A camp offers 4 different activities for an elective: archery, hiking, crafts and swimming. The capacity in each activity is limited so that at most 35 kids can do archery, 20 can do hiking, 25 can do crafts and 20 can do swimming. There are 100 kids in the camp. How many ways are there to assign the kids to the activities?

(4) A school cook plans her calendar for the month of February in which there are 20 school days. She plans exactly one meal per school day. Unfortunately, she only knows how to cook ten different meals. How many ways are there for her to plan her schedule of menus for the 20 school days if there are no restrictions on the number of times she cooks a particular type of meal?

Houseflies have short lifespans. Males of a certain species have lifespans that are strongly skewed to the right with a mean of 26 days and a standard deviation of 12 days. a) Explain why you cannot determine the probability that a given male housefly will live less than 24 days. b) Can you estimate the probability that the mean lifespan for a sample of 5 male houseflies is less than 24 days? Explain. c) A biologist collects a random sample of 65 of these male houseflies and observes them to calculate the sample mean lifespan. Describe the sampling distribution of the mean lifespan for samples of size 65. d) What is the probability that the mean lifespan for the sample of 65 houseflies is less than 24 days? e) What is the mean lifespan for the top 15% of samples of size 65?

Sample 1	Sample 2
12.1	8.9
9.5	10.9
7.3	11.2
10.2	10.6
8.9	9.8
9.8	9.8
7.2	11.2
10.2	12.1

A.

Construct the relevant hypotheses to test if the mean of the second population is greater than the mean of the first population.

B. Calculate the value of the test statistic. (Negative values should be indicated by a minus sign. Round all intermediate calculations to at least 4 decimal places and final answer to 3 decimal places.)

B-2. Find the p-value.

B-3. Do you reject the null hypothesis at the 1% level?

C. Do you reject the null hypothesis at the 10% level?

The average salary for American college graduates is $42,900. You suspect that the average is less for graduates from your college. The 49 randomly selected graduates from your college had an average salary of $38,173 and a standard deviation of $10,190. What can be concluded at the αα = 0.01 level of significance?

1. For this study, we should use Select an answert-test for a population meanz-test for a population proportion
2. The null and alternative hypotheses would be:

H0:H0:   ?pμ ?=≠><   

H1:H1:   ?pμ ?><≠=

3. The test statistic ?tz = (please show your answer to 3 decimal places.)
4. The p-value = (Please show your answer to 4 decimal places.)
5. The p-value is ?>≤ αα
6. Based on this, we should Select an answeracceptfail to rejectreject the null hypothesis.
7. Thus, the final conclusion is that ...
   • The data suggest that the populaton mean is significantly less than 42,900 at αα = 0.01, so there is statistically significant evidence to conclude that the population mean salary for graduates from your college is less than 42,900.
   • The data suggest that the population mean is not significantly less than 42,900 at αα = 0.01, so there is statistically insignificant evidence to conclude that the population mean salary for graduates from your college is less than 42,900.
   • The data suggest that the sample mean is not significantly less than 42,900 at αα = 0.01, so there is statistically insignificant evidence to conclude that the sample mean salary for graduates from your college is less than 38,173.
8. Interpret the p-value in the context of the study.
   • If the population mean salary for graduates from your college is $42,900 and if another 49 graduates from your college are surveyed then there would be a 0.10651237% chance that the population mean salary for graduates from your college would be less than $42,900.
   • There is a 0.10651237% chance that the population mean salary for graduates from your college is less than $42,900.
   • If the population mean salary for graduates from your college is $42,900 and if another 49 graduates from your college are surveyed then there would be a 0.10651237% chance that the sample mean for these 49 graduates from your college would be less than $38,173.
   • There is a 0.10651237% chance of a Type I error.
9. Interpret the level of significance in the context of the study.
   • There is a 1% chance that your won't graduate, so what's the point?
   • There is a 1% chance that the population mean salary for graduates from your college is less than $42,900.
   • If the population mean salary for graduates from your college is $42,900 and if another 49 graduates from your college are surveyed then there would be a 1% chance that we would end up falsely concluding that the population mean salary for graduates from your college is less than $42,900.
   • If the population population mean salary for graduates from your college is less than $42,900 and if another 49 graduates from your college are surveyed then there would be a 1% chance that we would end up falsely concluding that the population mean salary for graduates from your college is equal to $42,900.

The average amount of time it takes for couples to further communicate with each other after their first date has ended is 3.19 days. Is this average shorter for blind dates? A researcher interviewed 58 couples who had recently been on blind dates and found that they averaged 3 days to communicate with each other after the date was over. Their standard deviation was 0.639 days. What can be concluded at the  αα = 0.01 level of significance?

1. For this study, we should use Select an answer t-test for a population mean z-test for a population proportion
2. The null and alternative hypotheses would be:

H0:H0:   ?μp ?><≠=   

H1:H1:   ?pμ ?≠=<>

3. The test statistic ? tz = (please show your answer to 3 decimal places.)
4. The p-value = (Please show your answer to 4 decimal places.)
5. The p-value is ?>≤ αα
6. Based on this, we should Select an answerfail to reject accept reject the null hypothesis.
7. Thus, the final conclusion is that
   • The data suggest that the population mean is not significantly less than 3.19 at αα = 0.01, so there is statistically insignificant evidence to conclude that the population mean time for couples who have been on a blind date to communicate with each other after the date is over is less than 3.19.
   • The data suggest the populaton mean is significantly less than 3.19 at αα = 0.01, so there is statistically significant evidence to conclude that the population mean time for couples who have been on a blind date to communicate with each other after the date is over is less than 3.19.
   • The data suggest the population mean is not significantly less than 3.19 at αα = 0.01, so there is statistically significant evidence to conclude that the population mean time for couples who have been on a blind date to communicate with each other after the date is over is equal to 3.19.

Of the following, which is a reason we take samples instead of taking a census?

		Census is a hard word to say.
		A census is not reliable.
		A census doesn't give a good understanding of a whole group of people.
		A census is almost impossible to perform.
		All of the above.

A manufacturer produces both a deluxe and a standard model of an automatic sander designed for home use. Selling prices obtained from a sample of retail outlets follow.

	Model Price ($)				Model Price ($)
Retail Outlet	Deluxe	Standard		Retail Outlet	Deluxe	Standard
1	40	27		5	40	30
2	39	28		6	39	32
3	43	35		7	36	29
4	38	31

The manufacturer's suggested retail prices for the two models show a $10 price differential. Use a .05 level of significance and test that the mean difference between the prices of the two models is $10.

Develop the null and alternative hypotheses.
H 0 =  d Selectgreater than 10greater than or equal to 10equal to 10less than or equal to 10less than 10not equal to 10Item 1  
H a =  d Selectgreater than 10greater than or equal to 10equal to 10less than or equal to 10less than 10not equal to 10Item 2  

Calculate the value of the test statistic. If required enter negative values as negative numbers. (to 2 decimals).
  

The p-value is Selectless than .01between .10 and .05between .05 and .10between .10 and .20between .20 and .40greater than .40Item 4  

Can you conclude that the price differential is not equal to $10?
SelectYesNoItem 5  

What is the 95% confidence interval for the difference between the mean prices of the two models (to 2 decimals)? Use z-table.
( ,  )

3.3) A small market orders copies of a certain magazine for its magazine rack each week. Let X = demand for the magazine, with the following pmf.

x	1	2	3	4	5	6
p(x)	1 13	1 13	3 13	4 13	2 13	2 13

Suppose the store owner actually pays $2.00 for each copy of the magazine and the price to customers is $4.00. If magazines left at the end of the week have no salvage value, is it better to order three or four copies of the magazine? [Hint: For both three and four copies ordered, express net revenue as a function of demand X, and then compute the expected revenue.]

*What is the expected profit if three magazines are ordered? (Round your answer to two decimal places.)

*What is the expected profit if four magazines are ordered? (Round your answer to two decimal places.)

*How many magazines should the store owner order?

-3 magazines

-4 magazines

Based on historical data, your manager believes that 33% of the company's orders come from first-time customers. A random sample of 102 orders will be used to estimate the proportion of first-time-customers. What is the probability that the sample proportion is less than 0.27? Note: You should carefully round any z-values you calculate to 4 decimal places.

Answer =

(Enter your answer as a number accurate to 4 decimal places.)