Question

3. The Food Marketing Institute shows that 17% of households spend more than $100 per week on groceries. Assume the population proportion is p = .17 and a sample of 800 households will be selected from the population. a. Show the sampling distribution of p, the sample proportion of households spending more than $100 per week on groceries. b. What is the probability that the sample proportion will be within ±.02 of the population proportion? c. Answer part (b) for a sample of 1600 household

Answer 1

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 = $\sqrt{\frac{p(1-p)}{n} }$ = 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 = $\frac{P - p}{standard deviation}$

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 = $\frac{P-p}{standard deviation}$ 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

Answer 2

Answer:

See the steps below.

Step-by-step explanation:

a. Show the sampling distribution of p, the sample proportion of households spending more than $100 per week on groceries.  

The sample means = 0.17

The standard deviation of the sample means = $\sqrt{0.17*(0.83/800)}$ = 0.0133

Therefore, the sample proportion of households spending more than $100 per week on groceries is approximately 0.0133.

b) What is the probability that the sample proportion will be within plus or minus .02 of the population proportion?

Since the sample means = 0.17

For plus 0.02

The sample means = 0.17 + 0.02 = 0.19

Therefore, we have:

z(0.19) = $(0.19 - 0.17)/\sqrt{0.17*(0.83/800)}$ = 1.5060

For minus 0.02

The sample means = 0.17 - 0.02 = 0.15

z(0.15) = $(0.15 - 0.17)/\sqrt{0.17*(0.83/800) }$ = - 1.5060

Therefore, we now have:

P(0.15< p < 0.19) = P(-1.5060< z <1.5060) = 0.8679

c) Answer part (b) for a sample of 1600.

For plus 0.02

The sample means = 0.17 + 0.02 = 0.19

Therefore, we have:

z(0.19) = $(0.19 - 0.17)/\sqrt{0.17*(0.83/1,600)}$ = 2.1297

For minus 0.02

The sample means = 0.17 - 0.02 = 0.15

z(0.15) =$(0.15 - 0.17)/\sqrt{0.17*(0.83/1,600)}$ = - 2.1297

Therefore, we now have:

P(0.15< p < 0.19) = P(-2.197< z <2.1297) = 0.9660