are you a chocolate "purist", or do you like other ingredients in your chocolate? american demographics reports that almost 75% of consumers like traditional ingredients such as nuts or caramel in their chocolate. they are less enthusiastic about the taste of mint or coffee that provide more distinctive flavors. a random sample of 200 consumers is selected and the number who likes nuts or caramel in their chocolate is recorded.

(a) what is the approx. sampling distribution for the sample proportion p? what are the mean and standard deviation for this distribution?
(b) what is the probability that the sample percentage is greater than 80%?
(c) within what limits would you expect the sample proportion to lie about 95% of the time?

OUATTARA MAMANDOU

CCNY-CUNY
EXO: 7.47 page 281

P=0.75 Q=1-0.75 = 0.25 N=200
A- SE(^P)=sqrt (P.Q)/N where Q=1-P
SE(^P)=sqrt(0.75*0.25)/200
=0.306 SE(^P)=0.0306
Approximately normal with mean of 0.75 and standard deviation of 0.0306
B- Let find the Zscore of 80% z=^P-P/stddev
P(^P>80%) or P(^P>0.80)= P(0.80-0.75/0.0306)=1.63 ¡ú zvalue of 1.63 from the table is 0.9484 or P(z¡Ü1.63)=0.9484
Since P (^P>0.80)= 1-P(z¡Ü80). Then we compute 1-0.9484=0.0516.
So the probability that the sample proportion is greater than 80% is 0.0516
C- From the Empirical Rule and the general properties of the normal distribution, approximately 95% of the measurements will lie within 2 or 1.96 standard deviations of the mean:
P+-2SE since P=0.75 and SE(^P)=0.0306 we have 0.75+-2(0.0306)¡ú¡ú 0.75+-0.06
OR 0.75+0.06=0.81 and 0.75-0.06=0.69
Hence 95% lie between 0.69 and 0.81 OR 0.75+-0.06

Good job Mr Ouattara. Your answers are just excellent.

You are brillant.

To answer these questions, we will make use of the formulas for the mean and standard deviation of a sampling distribution for a proportion, as well as the concept of a confidence interval.

(a) The sampling distribution for the sample proportion p can be approximated using the normal distribution when the sample size is large enough. The mean of this distribution is the same as the true population proportion: 0.75, based on the given information. The standard deviation of the sampling distribution for the sample proportion can be calculated using the following formula:
σ = sqrt( (p * (1-p)) / n ), where p is the population proportion (0.75) and n is the sample size (200).
So, σ = sqrt( (0.75 * (1-0.75)) / 200) = sqrt( (0.1875) / 200) = sqrt(0.0009375) ≈ 0.0306.

(b) To calculate the probability that the sample percentage is greater than 80%, we need to convert this percentage to a proportion. The sample proportion is 0.80. We can then use the z-score formula to find the probability associated with this proportion. The z-score is calculated as:
z = (x - μ) / σ, where x is the sample proportion, μ is the population proportion (0.75), and σ is the standard deviation of the sampling distribution (0.0306).
Substituting the values, we get: z = (0.80 - 0.75) / 0.0306 ≈ 1.64.
Using a standard normal distribution table or a calculator, we can find the probability associated with this z-score. The probability that the sample percentage is greater than 80% is approximately 0.0495.

(c) To calculate the limits within which we can expect the sample proportion to lie about 95% of the time, we can use the concept of a confidence interval. A 95% confidence interval can be calculated as:
p ± z * sqrt( (p * (1-p)) / n), where p is the sample proportion, z is the critical value corresponding to the desired confidence level (in this case, 95%), and n is the sample size.
Substituting the values, the limits are given by:
0.75 ± (1.96 * sqrt( (0.75 * (1-0.75)) / 200 )).
Calculating this expression, we get the lower limit as 0.705 and the upper limit as 0.795.
Therefore, we would expect the sample proportion to lie within the range of approximately 0.705 to 0.795 about 95% of the time.

To find the answers to the questions, we can use the concepts of sampling distribution and binomial distribution.

(a) The sampling distribution for the sample proportion p can be approximated by a normal distribution when the sample size is large enough (n > 30). In this case, since we have a random sample of 200 consumers, this condition is satisfied.

The mean of the sampling distribution, denoted by μp, is equal to the population proportion. In this case, the population proportion is given as almost 75%, which can be expressed as 0.75.

The standard deviation of the sampling distribution, denoted by σp, can be calculated using the formula:

σp = sqrt[(p * (1 - p)) / n]

where p is the population proportion and n is the sample size.

Let's calculate the mean and standard deviation for this distribution:
μp = 0.75
σp = sqrt[(0.75 * (1 - 0.75)) / 200]

(b) To find the probability that the sample percentage is greater than 80%, we need to calculate the z-score and find the corresponding area under the normal distribution curve.

The formula for calculating the z-score is:
z = (x - μp) / σp

Here, x = 0.80 is the sample percentage we are interested in. Plug in the values of μp and σp from part (a) to calculate the z-score.

Then, using a standard normal distribution table or a calculator, find the probability corresponding to the z-score. Subtract this probability from 1 to get the probability that the sample percentage is greater than 80%.

(c) To find the limits within which we can expect the sample proportion to lie about 95% of the time, we use the concept of confidence intervals. A 95% confidence interval is commonly used, which means that we are 95% confident that the true population proportion falls within this interval.

The formula for calculating the confidence interval is:
CI = p̂ ± z * sqrt[(p̂ * (1 - p̂)) / n]

where p̂ is the sample proportion, n is the sample size, and z is the critical value corresponding to the desired confidence level. For a 95% confidence interval, the critical value is approximately 1.96.

Calculate the lower and upper limits for the confidence interval using the formula.

Now, let's calculate the answers to the questions using the given information and formulas.