Inactive Tutor answered 02/17/24
32 people = N, normal distribution, mean=43, s = 10
95% CI, z= 1.96= about 2
(x-43)/10/sqr32 = 2
20 = sqr32(x-3)
x = 3 + 20/sqr32 = 3 + 5/sqr2 = about 6.5355= $6.54
MOE = zs/sqrN = 20/sqr32 = 20/4sqr2 = 5/sqr2
Sophia R.
asked 12/02/22In a survey, 32 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $43 and standard deviation of $10. Find the margin of error at a 95% confidence level.
Inactive Tutor answered 02/17/24
32 people = N, normal distribution, mean=43, s = 10
95% CI, z= 1.96= about 2
(x-43)/10/sqr32 = 2
20 = sqr32(x-3)
x = 3 + 20/sqr32 = 3 + 5/sqr2 = about 6.5355= $6.54
MOE = zs/sqrN = 20/sqr32 = 20/4sqr2 = 5/sqr2
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