Question

The Food Marketing Institute shows that 15% of households spend
more than $100 per week on groceries. Assume the population
proportion is *p* = 0.15 and a sample of 800 households will
be selected from the population. Use *z*-table.

What is the probability that the sample proportion will be within +/- 0.02 of the population proportion (to 4 decimals)?

What is the probability that the sample proportion will be
within +/- 0.02 of the population proportion for a sample of 1,400
households (to 4 decimals)?

Answer #1

(1) What is the probability that the sample proportion will be
within +/- 0.02 of the population proportion (to 4 decimals)

sample proportion will be within +/- 0.02 means 0.15 + or -
0.02

or we need to estimate z(0.17) and z(0.13)

z(0.17)= difference / standard error = (0.17-0.15) / squrt [
0.15*0.85/800] = 1.5873

z(0.13) = - 1.5873

(2) What is the probability that the sample proportion will be
within +/- 0.02 of the population proportion for a sample of 1,400
households (to 4 decimals)

the new sample size is 1400

standard error of the proportion of households spending = sqrt [ p
x q / n ]

= sqrt [ 0.15*0.85/1400] = 0.0095

sample proportion will be within +/- 0.02 means 0.15 + or -
0.02

or we need to estimate z(0.17) and z(0.13)

z(0.17)= difference / standard error = (0.17-0.15) / squrt [
0.15*0.85/1400] = 2.1053

z(0.13) = - 2.1053

The Food Marketing Institute shows that 15% of households spend
more than $100 per week on groceries. Assume the population
proportion is p = 0.15 and a sample of 800 households will be
selected from the population. Use z-table.
What is the probability that the sample proportion will be
within +/- 0.02 of the population proportion for a sample of 1,800
households (to 4 decimals)?

The Food Marketing Institute shows
that 15% of households spend more than $100 per week on groceries.
Assume the population proportion is p = 0.15 and a sample
of 800 households will be selected from the population. Use
z-table.
Calculate (), the standard error of the proportion of households
spending more than $100 per week on groceries (to 4
decimals).
What is the probability that the sample proportion will be
within +/- 0.02 of the population proportion (to 4 decimals)?...

The Food Marketing Institute shows that 15% of households spend
more than $100 per week on groceries. Assume the population
proportion is p = 0.15 and a sample of 700 households will
be selected from the population. Use z-table.
Calculate (), the standard error of the proportion of
households spending more than $100 per week on groceries (to 4
decimals).
What is the probability that the sample proportion will be
within +/- 0.02 of the population proportion (to 4...

The Food Marketing Institute shows that 15% of households spend
more than $100 per week on groceries. Assume the population
proportion is p = 0.15 and a sample of 600 households will be
selected from the population. Use z-table.
Calculate (), the standard error of the proportion of households
spending more than $100 per week on groceries (to 4 decimals).
What is the probability that the sample proportion will be
within +/- 0.02 of the population proportion (to 4 decimals)?...

The Food Marketing Institute shows that 16% of households spend
more than $100 per week on groceries. Assume the population
proportion is p = 0.16 and a sample of 600 households will be
selected from the population. Use z-table.
Calculate (), the standard error of the proportion of households
spending more than $100 per week on groceries (to 4 decimals).
What is the probability that the sample proportion will be
within +/- 0.02 of the population proportion (to 4 decimals)?...

The Food Marketing Institute shows that 17% of households spend
more than $100 per week on groceries. Assume the population
proportion is p = 0.17 and a sample of 600 households will be
selected from the population. Use z-table.
What is the probability that the sample proportion will be
within +/- 0.02 of the population proportion for a sample of 1,200
households (to 4 decimals)?

The Food Marketing Institute shows that 18% of households spend
more than $100 per week on groceries. Assume the population
proportion is p = 0.18 and a sample of 700 households will
be selected from the population. Use z-table.
Calculate (), the standard error of the proportion of
households spending more than $100 per week on groceries (to 4
decimals).
What is the probability that the sample proportion will be
within +/- 0.03 of the population proportion (to 4...

The Food Marketing Institute shows that 16% of households spend
more than $100 per week on groceries. Assume the population
proportion is p = 0.16 and a sample of 900 households will be
selected from the population. Use z-table. Calculate (), the
standard error of the proportion of households spending more than
$100 per week on groceries (to 4 decimals). 0.0130 What is the
probability that the sample proportion will be within +/- 0.03 of
the population proportion (to 4...

The Food Marketing Institute shows that 17% of households spend
more than $100 per week on groceries. Assume the population
proportion is p = 0.17 and a sample of 600 households will be
selected from the population. Use z-table.
A) Calculate the standard error of the proportion of households
spending more than $100 per week on groceries (to 4 decimals).
I got .0153
B) What is the probability that the sample proportion will be
within +/- 0.03 of the population...

The Food Marketing Institute shows that 17% of households spend
more than $100 per week on groceries. Assume the population
proportion is
p = 0.17 and a sample of 600 households will
be selected from the population.
(b)What is the probability that the sample proportion will be
within ±0.02 of the population proportion? (Round your answer to
four decimal places.)
.8064
(c)Answer part (b) for a sample of 1,200 households. (Round your
answer to four decimal places.)

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