Question

In: Statistics and Probability

Germany introduced its version of the above game " Heimfreiheit und Spaß." The German branch strongly...

Germany introduced its version of the above game " Heimfreiheit und Spaß." The German branch strongly argued against using the standard deviation of the online game population. They argued that the game is unique in a unique environment. They reported the following daily revenue. Create a 92% confidence interval for the population mean of daily revenue in Germany Day Revenue

Day Revenue
1 $      5,756.67
2 $      9,830.94
3 $      4,816.01
4 $    14,223.89
5 $    10,165.92
6 $    11,536.27
7 $          369.86
8 $      6,653.34
9 $      4,094.15
10 $      8,991.33
11 $    18,661.26
12 $    19,761.52
13 $      9,941.33
14 $      4,562.90
15 $      5,048.30
16 $    10,797.53
17 $      2,095.75
18 $      7,080.88
19 $    11,508.74
20 $    20,999.13
21 $    13,782.45
22 $      6,777.79
23 $    13,548.91
24 $      2,302.33
25 $      8,151.19
26 $      9,048.90
27 $      8,723.91
28 $    20,045.47
29 $    11,861.94
30 $      9,267.34
31 $          125.79
32 $    11,564.47
33 $      9,663.64
34 $    10,827.95
35 $    13,924.31
36 $    20,185.78
37 $    20,882.18
38 $    10,009.74
39 $    10,734.91
40 $    18,305.92
41 $    13,791.64
42 $      1,195.78
43 $      9,118.23
44 $      8,062.51

Solutions

Expert Solution

They given reported daily revenue.

In order to find 92% daily confidence interval for population means of daily revenue in Germany day revenue.

The formula we have to use is following

Here n=44

Now to find mean and Standard deviation

100(1-alpha)%=92%

100(1-alpha)=92

1-alpha=0.92

alpha=1-0.92

alpha=0.08

My test is two-sided so alpha/2=0.08/2=0.04

Now see the value of 0.04 in the table and collect the corresponding value of Z(alpha/2)

Z(alpha/2)=1.75

Now to find X(bar)=10336

and Standard Deviation=5735

here n=44

Thus Using the formula of 92% Confidence interval for population means

The lower limit is 8822.978 and

The upper limit is 11849.02

CI is (8822.978, 11849.02)

Now When Population Standard Deviation is Unknown in this case we use the following formula to calculate CI.

Here Df=n-1=44-1=43

and Value of t is 1.793054 calculated by the command in excel =TINV(0.08,43)

Hence Lower limit is 7785.754

Upper Limit is 11886.25

Hence 92% CI is (7785.754, 11886.25)

Obs no Observations
1 5756.67 Mean 10336
2 9830.94 Standard Deviation 5735
3 4816.01 n 44
4 14223.89
5 10165.92 Z(alpha/2) 1.75
6 11536.27
7 369.86 Upper Limit 11849.02
8 6653.34 Lower Limit 8822.978
9 4094.15
10 8991.33
11 18661.26
12 19761.52
13 9941.33
14 4562.9
15 5048.3
16 10797.53
17 2095.75
18 7080.88
19 11508.74
20 20999.13
21 13782.45
22 6777.79
23 13548.91
24 2302.33
25 8151.19
26 9048.9
27 8723.91
28 20045.47
29 11861.94
30 9267.34
31 125.79
32 11564.47
33 9663.64
34 10827.95
35 19924.31
36 20185.78
37 20882.18
38 10009.74
39 10734.91
40 18305.92
41 13791.64
42 1195.78
43 9118.23
44 8062.51

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