Question

On May 3, 2017, the Calgary Herald reported that, according to Avalanche Canada, 24 of 45 people killed in avalanches over the previous 5 years were snowmobilers. Assume this is a random sample. a) Is the sample large enough that we can use techniques based on the normal distribution to analyze these data? Please explain. b) Calculate a 95% confidence interval for the proportion of avalanche deaths that are snowmobilers. c) Would a 99% confidence interval be wider or narrower? d) The claim is made that no more than 30% of all avalanche deaths are snowmobilers. What would the null and alternative hypotheses be if you want to show that the proportion is higher than 30% e) Based on your confidence interval, do you believe that more than 30% of all avalanche deaths are snowmobilers? Please explain.

Answer 1

Number of Items of Interest,   x =   24
Sample Size,   n =    45
      
Sample Proportion ,    p̂ = x/n =    0.5333

a)

np>5 and n(1-p)>5

here , np = 2

n (1-p) = 21

so,
sample is large enough that we can use techniques based on the normal distribution to analyze these data

.............

b)

z -value =   Zα/2 =    1.960   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0744          
margin of error , E = Z*SE =    1.960   *   0.0744   =   0.1458
                  
95%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.533   -   0.1458   =   0.3876
Interval Upper Limit = p̂ + E =   0.533   +   0.1458   =   0.6791
                  
95%   confidence interval is (   0.3876   < p <    0.6791   )

...............

c)

99%   confidence interval is (   0.3418   < p <    0.7249   )
99% CI is wider than 95%

............

d)

Ho :   p =    0.3  
H1 :   p >   0.3  

e)

0.3 does not lie in the CI so, reject Ho

so, it is concluded that more than 30% of all avalanche deaths are snowmobilers.

...............

Please revert back in case of any doubt.

Please upvote. Thanks in advance.