In 15 packages of \"Doggy Meat Snacks\" there were 48 lamb snacks. The total num

Question

In 15 packages of "Doggy Meat Snacks" there were 48 lamb snacks. The total number of snacks in the 15 bags was 120. We wish to calculate a 90% confidence interval for the population proportion of lamb snacks. Which distribution should you use for this problem? The sample proportion is The critical value z_crit of the standard normal distribution for the 90% confidence interval is The standard error for p is The error bound for the proportion (EBP) is The confidence interval is given by (L, U), where

Explanation / Answer

a)

We use a normal distribution for this problem.

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b)

There are a total of 48 lamb snacks out of 120 snacks.

Thus,

p^ = 48/120 = 0.4 [ANSWER]

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c)

Now, for the critical z,              
alpha/2 =   0.05          
Thus, z(alpha/2) =    1.645   [ANSWER]      
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d)              

Also, we get the standard error of p, sp:              
sp = sqrt[p^ (1 - p^) / n] =    0.04472136   [ANSWER]

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e)
              
Thus,              

Margin of error = z(alpha/2)*sp =    0.073566636 [ANSWER]

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f)
          
lower bound = p^ - z(alpha/2) * sp =   0.326433364 [ANSWER]          
upper bound = p^ + z(alpha/2) * sp =    0.473566636   [ANSWER]      
              

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