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Acme Confectionery has launched a new bar called Chocolate Crunch - Leaving Cert Mathematics - Question 8 - 2018

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Acme Confectionery has launched a new bar called Chocolate Crunch. The weights of these new bars are normally distributed with a mean of 64.6 g and a standard deviat... show full transcript

Worked Solution & Example Answer:Acme Confectionery has launched a new bar called Chocolate Crunch - Leaving Cert Mathematics - Question 8 - 2018

Step 1

Find the probability that the mean weight of the sample is between 4.6 g and 4.7 g.

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Answer

To find this probability, we will use the Z-score formula:

Z=Xμσ/nZ = \frac{X - \mu}{\sigma / \sqrt{n}}

For the given values:

  • Mean ((\mu)) = 64.6 g
  • Standard deviation ((\sigma)) = 0.12 g
  • Sample size (n) = 10
  1. Calculating the Z-score for 4.6 g:

    • (Z_1 = \frac{4.6 - 64.6}{0.12 / \sqrt{10}} = \frac{-60}{0.0379} = -1581.38)
  2. Calculating the Z-score for 4.7 g:

    • (Z_2 = \frac{4.7 - 64.6}{0.12 / \sqrt{10}} = \frac{-59.9}{0.0379} = -1580.66)
  3. Use standard normal distribution tables or calculators to find probabilities. The probabilities are insignificantly small, leading to:

    • The probability that the mean is between 4.6 g and 4.7 g approaches 0.

Step 2

Create the 95% confidence interval for the population proportion who liked the new bar.

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Answer

The sample proportion (\hat{p}) is given by:

p^=324400=0.81\hat{p} = \frac{324}{400} = 0.81

Next, calculate the standard error (SE):

SE=p^(1p^)n=0.81(10.81)400=0.810.194000.025SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.81(1 - 0.81)}{400}} = \sqrt{\frac{0.81 \cdot 0.19}{400}} \approx 0.025

Now, for a 95% confidence interval, the Z-score is approximately 1.96. Thus, the margin of error (ME) is:

ME=ZimesSE=1.96×0.0250.049ME = Z imes SE = 1.96 \times 0.025 \approx 0.049

Now we find the confidence interval:

0.810.049p0.81+0.0490.81 - 0.049 \leq p \leq 0.81 + 0.049

So,

0.76p0.860.76 \leq p \leq 0.86

The 95% confidence interval for the population proportion who liked the new bar is (0.76, 0.86).

Step 3

Put one tick into the table for each statement.

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Answer

  1. When forming confidence intervals (for fixed n and (\hat{p})), an increased confidence level implies a wider interval. -> Always True.

  2. As the value of (\hat{p}) increases (for fixed n), the estimated standard error of the population proportion increases. -> Sometimes True.

  3. As the value of (\hat{p}(1 - \hat{p})) increases (for fixed n), the estimated standard error of the population proportion increases. -> Sometimes True.

  4. As n, the number of people sampled, increases (for fixed (\hat{p})), the estimated standard error of the population proportion increases. -> Never True.

Step 4

Using calculus or otherwise, find the maximum value of \(\hat{p}(1 - \hat{p})\).

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Answer

We can express the function as:

M=p^(1p^)=p^p^2M = \hat{p}(1 - \hat{p}) = \hat{p} - \hat{p}^2

To find the maximum value, we differentiate:

dMdp^=12p^=0\frac{dM}{d\hat{p}} = 1 - 2\hat{p} = 0

Solving for (\hat{p}):

p^=12\hat{p} = \frac{1}{2}

Now, substituting back into the original equation:

Mmax=12(112)=14=0.25M_{max} = \frac{1}{2} \left( 1 - \frac{1}{2} \right) = \frac{1}{4} = 0.25

Thus, the maximum value of (\hat{p}(1 - \hat{p})) is 0.25.

Step 5

Find the largest possible value of the radius of the 95% confidence interval.

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Answer

Using the result from the previous part, we substitute into:

R=1.960.25nR = 1.96 \sqrt{\frac{0.25}{n}}

For a sample size of 800:

R=1.960.25800=1.96×0.01770.0347R = 1.96 \sqrt{\frac{0.25}{800}} = 1.96 \times 0.0177 \approx 0.0347

Hence, the largest possible radius of the 95% confidence interval for a population proportion, given a random sample of size 800, is approximately 0.0347.

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