Question

MNM Corporation gives each of its 500 employees an aptitude test. The scores on the test are normally distributed with a mean of 75 and a standard deviation of 15. A simple random sample of 36 is taken from the population of 500 employees. What is the probability that the average aptitude test score in the sample will be less than 78.69? A new brand of breakfast cereal is being market tested. One hundred boxes of the cereal were given to consumers to try. The consumers were asked whether they liked or disliked the cereal. Their responses are provided below. Construct a 95% confidence interval for the true proportion of people who will like the cereal. Interpret the practical meaning of this interval estimate, in plain English. Assume that nothing is known about the population proportion. With a .95 probability, how large of a sample needs to be taken to provide a margin of error of 9% or less?

Answer 1

Answer:

Q1: $P(\bar X$

Q2: $0.6 - 1.96 \sqrt{\frac{0.6(1-0.6)}{100}}=0.504$  

$0.6 + 1.96 \sqrt{\frac{0.6(1-0.6)}{100}}=0.696$  

And the 95% confidence interval would be given (0.504;0.696).

We are confident 95% that about 50.4% to 69.6% of people would like the cereal

Q3: $n=\frac{0.5(1-0.5)}{(\frac{0.09}{1.96})^2}=118.57$  

And rounded up we have that n=119  

Step-by-step explanation:

Q1 :MNM Corporation gives each of its 500 employees an aptitude test. The scores on the test are normally distributed with a mean of 75 and a standard deviation of 15. A simple random sample of 36 is taken from the population of 500 employees. What is the probability that the average aptitude test score in the sample will be less than 78.69?

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Let X the random variable that represent the scores of a population, and for this case we know the distribution for X is given by:

$X \sim N(75,15)$  

Where $\mu=75$ and $\sigma=15$

The distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu=75, \frac{\sigma}{\sqrt{n}}=\frac{15}{\sqrt{36}}=2.5)$

For this case we want this probability:

$P(\bar X$

And we can use the z score given by:

$z= \frac{\bar X -\mu}{\sigma_{\bar X}}$

$P(\bar X$

Q2: A new brand of breakfast cereal is being market tested. One hundred boxes of the cereal were given to consumers to try. The consumers were asked whether they liked or disliked the cereal. Their responses are provided below.

Liked =60, Disliked =40 , Total =100

Construct a 95% confidence interval for the true proportion of people who will like the cereal. Interpret the practical meaning of this interval estimate, in plain English.

n=100 represent the random sample taken      

X=60 represent the people liked with the cereal

$\hat p=\frac{60}{100}=0.6$ estimated proportion of people who liked the cereal

$\alpha=0.05$ represent the significance level (no given, but is assumed)      

Confidence =95% or 0.95  

p= population proportion of people drivers claimed they always buckle up

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

The confidence interval would be given by this formula  

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$  

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$, with that value we can find the quantile required for the interval in the normal standard distribution.  

$z_{\alpha/2}=1.96$  

And replacing into the confidence interval formula we got:  

$0.6 - 1.96 \sqrt{\frac{0.6(1-0.6)}{100}}=0.504$  

$0.6 + 1.96 \sqrt{\frac{0.6(1-0.6)}{100}}=0.696$  

And the 95% confidence interval would be given (0.504;0.696).

We are confident 95% that about 50.4% to 69.6% of people would like the cereal

Q3: Assume that nothing is known about the population proportion. With a .95 probability, how large of a sample needs to be taken to provide a margin of error of 9% or less?

The margin of error for the proportion interval is given by this formula:  

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)  

And on this case we have that $ME =\pm 0.09$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

We can assume that the estimated proportion is 0.5 since we don't have other info provided to assume a different value. And replacing into equation (b) the values from part a we got:  

$n=\frac{0.5(1-0.5)}{(\frac{0.09}{1.96})^2}=118.57$  

And rounded up we have that n=119