Question

In six packages of “The Flintstones® Real Fruit Snacks” there were five Bam-Bam snack pieces. The total number of snack pieces in the six bags was $68.$We wish to calculate a $96\%$confidence interval for the population proportion of Bam-Bam snack pieces.

a. Define the random variables $XandP\prime$in words.

b. Which distribution should you use for this problem? Explain your choice

c. Calculate$p\prime .$

d. Construct a $96\%$confidence interval for the population proportion of Bam-Bam snack pieces per bag.

i. State the confidence interval.

ii. Sketch the graph.

iii. Calculate the error bound.

e. Do you think that six packages of fruit snacks yield enough data to give accurate results? Why or why not?

Short answer

(a) The sample of $68$snack pieces for the proportion of Bam-Bam pieces is $P\text{'},$and the bag of snacks for the proportion of Bam-Bam snacks is X.

(b) From this we use normal distribution $N\left(P\sqrt{\frac{pq}{n}}\right)$.

(c) After calculate we get, $p^{\text{'}}=0.0735$.

(d) The final result

1. we get$CI=(0.00853,0.13853)$

2.

3. Error bond mean $EBM=0.0554$.

(e) Almost certainly not.

Step-by-step solution

Explanation (a)

The sample of $68$snack pieces for the proportion of Bam-Bam pieces is $P\text{'},$ and the bag of snacks for the proportion of Bam-Bam snacks is$X$.

Explanation (b)

The problem is solved using the normal distribution. As we all know, the sample size for the population's standard deviation is greater than 30, and the distribution is$N\left(P,\sqrt{\frac{pq}{n}}\right)$.

Explanation (c)

In this we calculate p

By using formula

$p^{\text{'}}=\frac{5}{\text{total number of smack pieces}}$

by Computing,

$p^{\text{'}}=\frac{5}{68}$

$p^{\text{'}}=0.0735$.

Explanation (d)

i. The confidence interval should be stated.

The confidence interval's output,

$n=58$

$CI=(0.00853,0.13853)$

ii. The graph is as follows:

iii. The formula is used to compute the error bound.

$EBM=z_{\frac{\alpha }{2}}\sqrt{\frac{p^{\text{'}}{q}^{\text{'}}}{n}}$

$EBM=\frac{20.04}{2}\sqrt{\frac{0.0735\times 0.9265}{68}}$

localid="1651847585804" $EBM=0.0554$

Explanation (e)

Do you believe that six packages of fruit snacks will provide enough data to provide accurate results, It's unlikely, because the samples were obtained using a basic random sample.