Question

____________________________________________

- The results of a sample of 372 subscribers to
*Wired*magazine shows the time spent using the Internet during the week. Previous surveys have revealed that the population standard deviation is 10.95 hours. The sample data can be found in the Excel test data file.

- What is the probability that another sample of 372 subscribers spends less than 19.00 hours per week using the Internet?

____________________________________________

- Develop a 95% confidence interval for the population mean

____________________________________________

- If the editors of
*Wired*wanted to have the error be no more than 1 hour, using the same 95% confidence interval, what size of sample would be required.

____________________________________________

35 |

14 |

7 |

24 |

2 |

22 |

14 |

15 |

36 |

13 |

37 |

25 |

33 |

14 |

11 |

8 |

7 |

15 |

19 |

6 |

3 |

27 |

29 |

35 |

6 |

32 |

10 |

5 |

2 |

12 |

21 |

9 |

38 |

32 |

24 |

10 |

15 |

25 |

10 |

7 |

32 |

12 |

18 |

26 |

35 |

38 |

37 |

11 |

25 |

24 |

32 |

35 |

11 |

31 |

4 |

25 |

15 |

13 |

37 |

5 |

21 |

36 |

34 |

22 |

13 |

12 |

12 |

38 |

23 |

13 |

15 |

11 |

10 |

19 |

9 |

22 |

29 |

17 |

26 |

36 |

35 |

3 |

14 |

32 |

20 |

13 |

23 |

3 |

10 |

15 |

32 |

8 |

21 |

31 |

31 |

8 |

5 |

17 |

18 |

28 |

19 |

8 |

33 |

7 |

34 |

23 |

27 |

4 |

5 |

23 |

37 |

34 |

25 |

21 |

37 |

27 |

4 |

6 |

21 |

28 |

24 |

12 |

21 |

3 |

10 |

20 |

14 |

11 |

18 |

10 |

21 |

5 |

25 |

22 |

2 |

10 |

4 |

4 |

17 |

8 |

29 |

7 |

5 |

5 |

38 |

14 |

35 |

13 |

16 |

17 |

13 |

17 |

31 |

2 |

13 |

8 |

11 |

27 |

32 |

18 |

20 |

25 |

3 |

22 |

19 |

7 |

3 |

35 |

12 |

4 |

30 |

30 |

15 |

23 |

3 |

31 |

35 |

23 |

4 |

34 |

7 |

24 |

17 |

29 |

3 |

37 |

30 |

10 |

28 |

7 |

17 |

35 |

5 |

4 |

12 |

2 |

5 |

31 |

9 |

30 |

10 |

9 |

20 |

7 |

22 |

18 |

24 |

7 |

6 |

5 |

16 |

19 |

19 |

32 |

14 |

5 |

14 |

34 |

3 |

29 |

9 |

33 |

8 |

35 |

3 |

20 |

4 |

13 |

13 |

8 |

30 |

23 |

27 |

2 |

36 |

35 |

12 |

35 |

6 |

10 |

23 |

29 |

2 |

34 |

31 |

13 |

19 |

12 |

38 |

15 |

35 |

30 |

22 |

10 |

20 |

16 |

15 |

38 |

21 |

35 |

7 |

2 |

8 |

10 |

15 |

28 |

24 |

14 |

18 |

25 |

38 |

28 |

33 |

2 |

4 |

4 |

15 |

19 |

12 |

25 |

20 |

29 |

22 |

24 |

15 |

3 |

5 |

37 |

33 |

22 |

18 |

19 |

37 |

16 |

29 |

11 |

9 |

3 |

5 |

5 |

18 |

25 |

18 |

14 |

20 |

37 |

10 |

14 |

3 |

3 |

12 |

3 |

30 |

5 |

30 |

15 |

22 |

29 |

36 |

14 |

25 |

3 |

22 |

26 |

7 |

22 |

26 |

18 |

36 |

33 |

3 |

23 |

5 |

28 |

7 |

28 |

6 |

8 |

17 |

25 |

20 |

4 |

22 |

25 |

31 |

18 |

32 |

21 |

35 |

35 |

29 |

27 |

2 |

5 |

21 |

26 |

26 |

5 |

14 |

17 |

7 |

19 |

11 |

7 |

20 |

36 |

7 |

20 |

32 |

23 |

32 |

2 |

Answer #1

Given,

Mean | 18.54301 |

Sd | 10.8051 |

n | 372 |

σ = 10.95

a)

P(X bar < 19)

Z = (X bar - Mean)/(σ/SQRT(n)) = (19-18.543)/(10.95/SQRT(372)) = 0.81

P(Z < 0.8050) = **0.7910**

b)

95% CI

Given | |

X bar | 18.543 |

S | 10.95 |

n | 372 |

α= | 0.05 |

CI | ||

Zc | 1.959964 | +/- NORM.S.INV(α/2) |

Upper | 19.65573 | X bar + Zc*(S/SQRT(n)) |

Lower | 17.43027 | X bar - Zc*(S/SQRT(n)) |

CI = (17.4303, 19.6557)

c)

σ = 10.95

ME = 1

C = 0.95

Zc = 1.96

n >= (Zc*σ/ME)^2 = (1.96*10.95/1)^2 = 460.62 = 461

The results of a sample of 372 subscribers
toWiredmagazine shows the time spent using the Internet
during the week. Previous surveys have revealed that the population
standard deviation is 10.95 hours. The sample data can be found in
the Excel test data file.
What is the probability that another sample of 372 subscribers
spends less than 19.00 hours per week using the Internet?
Develop a 95% confidence interval for the population mean
If the editors of Wiredwanted to have the...

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