Could you please work out this and show answers if possible :)
1 answer:
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Answer:
A)sample proportion = 0.17, the sampling distribution of p can be calculated/approximated with normal distribution of sample proportion = 0.17 and standard error/deviation = 0.013281
B) 0.869
C)0.9668
Step-by-step explanation:
A) p ( proportion of population that spends more than $100 per week) = 0.17
sample size (n)= 800
the sample proportion of p = 0.17
standard error of p = = 0.013281
the sampling distribution of p can be calculated/approximated with
normal distribution of sample proportion = 0.17 and standard error/deviation = 0.013281
B) probability that the sample proportion will be +-0.02 of the population proportion
= p (0.17 - 0.02 ≤ P ≤ 0.17 + 0.02 ) = p( 0.15 ≤ P ≤ 0.19)
z value corresponding to P
Z =
at P = 0.15
Z = (0.15 - 0.17) / 0.013281 = = -1.51
at P = 0.19
z = ( 0.19 - 0.17) / 0.013281 = 1.51
therefore the required probability will be
p( -1.5 ≤ z ≤ 1.5 ) = p(z ≤ 1.51 ) - p(z ≤ -1.51 )
= 0.9345 - 0.0655 = 0.869
C) for a sample (n ) = 1600
standard deviation/ error = 0.009391 (applying the equation for calculating standard error as seen in part A above)
therefore the required probability after applying
z = at p = 0.15 and p = 0.19
p ( -2.13 ≤ z ≤ 2.13 ) = p( z ≤ 2.13 ) - p( z ≤ -2.13 )
= 0.9834 - 0.0166 = 0.9668