Question

The published universal distribution of M&Ms by color is 13% brown, 13% red, 14% yellow, 24% blue, 20%

orange, and 16% green. You find a snack pack, which has about 20 candies; you open it, calculate the distribution,
and then eat them all. In that pack, 15% are orange. Curious, you find a single-serve bag, containing about
50 candies. In this bag, 24% are orange. More intrigued, you go for a king-size bag, which ends up containing
about 19% orange. You try some more bags, but none of the bags you try contain the actual published percentage
of orange M&Ms. Does this make you think that their figures are wrong and that they should publish the correct
percentages?

Answer 1

See explanation below.

We don't have enough evidence to conclude that the true proportion is different from 0.2 or 20% of 10% of significance.

Explanation:

Previous concepts

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.

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".

Solution to the problem

We can check if the statement is correct using a proportion interval, let's assume a significance level of 10% and a confidence of 90%.

The population proportion have the following distribution

`p \sim N(p,\sqrt{(p(1-p))/(n)})`

The confidence interval for a proportion is given by this formula

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

For the 90% confidence interval the value of
`\alpha=1-0.90=0.1` and
`\alpha/2=0.05`, with that value we can find the quantile required for the interval in the normal standard distribution.

`z_(\alpha/2)=1.64`

Snack pack

And replacing into the confidence interval formula we got:

`0.15 - 1.64 \sqrt{(0.15(1-0.15))/(20)}=0.019`

`0.15 + 1.64 \sqrt{(0.15(1-0.15))/(20)}=0.281`

And the 90% confidence interval would be given (0.019;0.281). So then we see that the claimed proportion is on the interval 20% or 0.2 so then we don't have problems with the claim on this case at the confidence level assumed

Single serve bag

And replacing into the confidence interval formula we got:

`0.24 - 1.64 \sqrt{(0.24(1-0.24))/(50)}=0.141`

`0.24 + 1.64 \sqrt{(0.24(1-0.24))/(50)}=0.339`

And the 90% confidence interval would be given (0.141;0.339). So then we see that the claimed proportion is on the interval 20% or 0.2 so then we don't have problems with the claim on this case at the confidence level assumed.

For the king size we don't have the sample size provided but on this case we can assume that if for the snack and the single pack we obtain that the proportion is not significant different from 0.2. Then is not correct say that the true proportion of orange candies is different from 0.2 or 20% at 10% of significance.