a grocery store sells snacks by weight .a six ounce bag of mixed nuts costs $3.60 . predict the cost of two ounce bag
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Answer:
a. The number of people that should be in the pilot study are 600 people
b. The point estimate is 0.62
c. At 95% confidence level the true population proportion of potential car buyers of hybrid vehicle is between the confidence interval (0.588, 0.6654)
d. Two ways to reduce the margin of error are;
1) Reduce the confidence interval
2) Use a larger sample size
Step-by-step explanation:
a. The given parameters for the estimation of sample size is given as follows;
The margin of error for the confidence interval, E = 4% = 0.04
The confidence level = 95%
The sample size formula for a proportion as obtained from an online source is given as follows;
Where, P is the estimated proportions of the desired statistic, therefore, we have for a new study, P = 0.5;
Z = The level of confidence at 95% = 1.96
n + The sample size
Therefore, we have;
Therefore, the number of people that should be in the pilot study in order to meet this goal at 95% confidence level is n = 600 people
b. The point estimate for the population proportion is the sample proportion given as follows;
Where;
x = The number of the statistic in the sample
n = The sample size
From the question, we have;
The number of potential car buyers, n = 600
The number of respondent in the sample that indicated that they would consider purchasing a hybrid, x = 376
Therefore, the point estimate, for the proportion of potential car buyers that would consider buying a hybrid vehicle, = 376/600 = 0.62
c. The confidence interval for a proportion is given as follows
Therefore, we get;
C.I. ≈ 0.6267 ± 0.0387
The 95% confidence interval for the true population proportion of potential buyers of hybrid vehicle, C.I. = (0.588, 0.6654)
d. The margin of error is given by the following formula;
Where;
= Margin of error at a given level of confidence
= z-score
σ = The standard deviation
n = The sample size
Therefore, the margin error can be reduced by the following two ways;
1) Reducing the confidence interval and therefore, the z-score
2) Increasing the sample size
We are looking for the probability :
Transform the law to standard normal like this:
The above formula is equivalent to this one:
From normal law table, we read the value of
.
Solving the above equation for the score n:
, it is the score we are looking for.
Transform the law to standard normal like this:
The above formula is equivalent to this one:
From normal law table, we read the value of
Solving the above equation for the score n: