Question

This is a big task. Need in 1hour 30min, ONLY NEED: 1D, 1E, 3B, 3C, 3D

I AssociatelDJuneCustr JunedSalesJurne$Sales Jul July Sales Reion JuneTraining? Evaluntions of Sales Associates at an Industrial Machinery Wholesaler 14 24y 1617236 AswcialelD Sales asciate's c June Custoners-Number ofcusnomers assoclae met wish duing Jane Jone#Sales-Number of sales placed by associate during June une SSales Total amunt, in doilars, ofassociae's sases during June uly Customers-Namber of cstomers associate met wish during Jaly July#Sales-Number of sales placed by associate during July ulySaies Total amount, in dollars, of associate's sales duing Juih RegionIocationef nplayee'sStae: North, Suth, Eat, or West June Training-Whether or not associate completed α ales raining program u! the end of June 59136 E 49226 W company idlentifhcativn n 41 16836 8380 1013830 34910 23 24754 44186 4383) S 42629 S 20958 E 23718 N 8354 WW 37241N 45 27190 E 26096 S 23168 H 29494 E 37871 N 33822 W 17121 H 15650 W 28065 S 51982 S 4*1 57608 2625372 1217894 14044 1590925 0.99999902 821996 14522 33385 3435 19 23806 36856 7528 37413 21 12285 12524 29325 33260 17198 25.3598231 523 37210 W 15218 N 32304 Ww 25226 S 50050 N 16279 S 38O9) 30 141 A population data set contains all ncmbers of a speciied group, in this sceaario we have 52 associates. A sample data set contains a part, or a subset 13525 E 16100 w 4180 S 16920 21.2176923 Mean .384709632 Standard Deviation 36 41 16 28208 37144 50450 38386 22474 E 19957 E 239K N 37345 W 28890 N 12970 E 1 240174973 Sandard Eror 0.624037721 1.026542051 Margin of Error 091 28140 29149 30 38 12 12 The mean of is probably around 21. I a90% conlident that this is rue. 24464 E 13446 N 29847 N would like to esmae apopulation proportion hecause I want tn know hea many associates had more than 25 29370 16 20442 19473 W 32070 N 27598 S 24186 N 34563 N

PART 1 - For now, you assume both of the claims about the population are correct.

1a. Given the assumed pop. mean and st.dev, calculate the probability of observing a value above the number for your first data point in your file.
1b. Suppose you collected 8 new data points in a new sample. Calculate the probability that the mean of these 8 new data points is above the number for your first data point in your file.
1c. If this is a normally distributed variable, above what value should you find 70% of data points? How many of the values from your data set are above this value?
1d. If this is a normally distributed variable, between what two numbers (centered around the assumed mean) should you find 68% of data points? What percentage of your data points are between these numbers?
1e. Think about your answers to 1c and 1d. Does this variable appear to be normally distributed with this mean and standard deviation?

PART 2 - Suppose you are now suspicious of both claims about the population.

2a. Describe the sample and population you are studying with this data.  (This is not given in the data, you have to make up a story.)
2b. Calculate the sample mean and sample standard deviation for your variable. Keep at least four decimal places
2c. Calculate the standard error and the margin of error for a 90% confidence interval. Keep at least four decimal places.
2d. Combine these three responses to calculate and interpret a 90% confidence interval for the population mean. Use the format: "The mean of <sample> is ###. Therefore I believe the mean of <population> is probably around ###. I am 90% confident that ______."

PART 3 - Create a confidence interval for a proportion.  
Drug Trial Data - Investigate the proportion of married patients.
Quality Control Data - Investigate the proportion of chips chosen while Kelly was managing the factory.
Employee Evaluation Data - Investigate the proportion of employees in the north region.

3a. Write a reason why you would want to estimate this population proportion.
3b. State the sample size, n, and the sample proportion, "p bar" (p with a line over it).
3c. Calculate a 95% confidence interval for the population proportion, p. Report at least three decimal places.
3d. Interpret this interval in the same way you did in step 2d.

Answer 1

a)

4.5 μ = 23, σ P(X > 32) =? X- 11 32-23 . P(z > 2) = 0.02275

b)

μ 23, σ 4.5, n = 8 P(X > 32) =? P(X > 32) = P(X-μ > 32-23) P(z > 5.65) = 0 一,1 ..

c)

11 23.ơ 4.5 P(X > x1) = 0.70.2.1 =? We know from standard normal table P(Z >-0.5244) = 0.70 1 -23 4.50.5244 r10.5244 * 4.52320.6402

So above 20.64, we will have 70% of the data.

We have 31 data points above this value, which is 57.4 %

d)

P(z! < X < x2) = 0.68,T1,T2 =? We know from standard normal table -0.9945 < Z < 0.9945) = 0.68 -23 =-0.9945 45 0.9945 ri =ー0.9945 * 4.5 T2 - 23 23-18.525 2 0.9945 4.5 r2 0.9945 * 4.5 + 23 = 27.475

We have 25 data points within this interval in our data set, which is 46.29%

e)

Given our values in the above two parts and the proportion of values that are in those intervals, its seems hard to believe that the data follows a normal distribution.

Below is the data of July $sales.

July$Sales
10
10
12
13
13
13
14
14
14
15
15
16
16
16
17
17
17
17
20
20
20
20
21
21
21
21
22
22
23
24
24
25
25
25
25
25
25
26
26
26
26
27
27
29
29
29
31
31
32
35
36
36
37