Main Answer: The standard error of the proportion of households spending more than $100 per week on groceries is approximately 0.013, rounded to four decimal places.The probability that the sample proportion will be within +/-0.03 of the population proportion is approximately 0.95, rounded to four decimal places.The probability that the sample proportion will be within +/-0.03 of the population proportion for a sample of 1,300 households is approximately 0.96, rounded to four decimal places.The option C is correct.
Step-by-step explanation:
In this exercise, we are given the population proportion p=0.16 and a sample size of n=900. We will use the z-table to calculate the standard error and probability of the sample proportion being within a certain range of the population proportion.
To calculate the standard error of the proportion, we use the formula:
σ(pˉ) = sqrt(p(1-p)/n)
Substituting p=0.16 and n=900 into this formula, we get:
σ(pˉ) = sqrt(0.16(1-0.16)/900) = 0.013, rounded to four decimal places.
To find the probability that the sample proportion will be within +/-0.03 of the population proportion, we use a z-score table to find the critical value for a confidence level of 95%. We then use this critical value to calculate the range within which we are confident that the sample proportion will fall with a certain level of confidence.
First, we find the critical value for a confidence level of 95% using a z-score table:z= 1.96 (rounded to four decimal places)Next, we calculate the range within which we are confident that the sample proportion will fall:
pˉ ± zσ(pˉ) = p ± zσ(pˉ) = p ± (zσ(pˉ))/n = p ± (1.96(0.013))/900 = p ± 0.03 (rounded to four decimal places)Therefore, with a confidence level of 95%, we are confident that the sample proportion will fall within +/-0.03 of the population proportion.
To find the probability that the sample proportion will be within +/-0.03 of the population proportion for a sample size of n=1,300, we repeat steps a and b above using n=1,300 instead of n=900:σ(pˉ) = sqrt(p(1-p)/n) = sqrt(0.16(1-0.16)/1300) = 0.012, rounded to four decimal places.
z= 1.96 (rounded to four decimal places). The range within which we are confident that the sample proportion will fall is: pˉ ± zσ(pˉ) = p ± (zσ(pˉ))/n = p ± (1.96(0.012))/1300 = p ± 0.03 (rounded to four decimal places).
Therefore, with a confidence level of 95%, we are confident that the sample proportion will fall within +/-0.03 of the population proportion for a sample size of n=1,30.The option C is correct.