Question

Chrome File Edit View History Bookmarks People Tab Window Help 60% Wed Jul 29 6:44 PM Home: S2020STAT205L01:0 x You are screen sharing Stop Share с webwork.ucalgary.ca/webwork2/S2020STAT205L01/ ra300166748 key XV862. To points) You are interested in determining what proportion of active NHL players are Canadian. Suppose a survey of 194 current active NHL players is taken. Of the players surveyed, 84 are Canadian Remaining time: 40:04 (min:sec) Use this information to answer the following questions. Please round your final answer to at least four decimats if necessary, a. Report the value of ø, the proportion of sampled players who are Canadian Answer: p= 0.4329 b. You wish to compute a 94% confidence interval for p, the proportion of all active NHL players who are Canadian. What is the value of the multiplier that you would use in the calculation of this interval? Please report the positive (absolute value) of the multiplier Answer: c. Computo a 94% confidence interval for p, the proportion of all active NHL players who are Canadian. Roport your lower bound in the first answer box and your upper bound in the 94% confidence interval second answer box d. Interpret the meaning of the above confidence interval We are 94% confident that the percent of ? who are Canadian is ? !!! % and %. JU 29 W P O ...

Answer 1

TRADITIONAL METHOD
given that,
possible chances (x)=84
sample size(n)=194
success rate ( p )= x/n = 0.433
I.
sample proportion = 0.433
standard error = Sqrt ( (0.433*0.567) /194) )
= 0.0356
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.06
from standard normal table, two tailed z α/2 =1.881
margin of error = 1.881 * 0.0356
= 0.0669
III.
CI = [ p ± margin of error ]
confidence interval = [0.433 ± 0.0669]
= [ 0.3661 , 0.4999]
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DIRECT METHOD
given that,
possible chances (x)=84
sample size(n)=194
success rate ( p )= x/n = 0.433
CI = confidence interval
confidence interval = [ 0.433 ± 1.881 * Sqrt ( (0.433*0.567) /194) ) ]
= [0.433 - 1.881 * Sqrt ( (0.433*0.567) /194) , 0.433 + 1.881 * Sqrt ( (0.433*0.567) /194) ]
= [0.3661 , 0.4999]
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interpretations:
1. We are 94% sure that the interval [ 0.3661 , 0.4999] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 94% of these intervals will contains the true population proportion
a.
proportion pf sampled players who are canadian = 0.433
b.
multiplier in the confidence interval is z α/2 =1.881
c.
94% sure that the interval [ 0.3661 , 0.4999]
d.
interpretations:
1. We are 94% sure that the interval [ 0.3661 , 0.4999] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 94% of these intervals will contains the true population proportion