Karen wants to advertise how many chocolate chips are in each Big Chip cookie at her bakery. She randomly selects a sample of 46 cookies and finds that the number of chocolate chips per cookie in the sample has a mean of 6.3 and a standard deviation of 3.1. What is the 99% confidence interval for the number of chocolate chips per cookie for Big Chip cookies? Enter your answers accurate to one decimal place (because the sample statistics are reported accurate to one decimal place).

1 answer

To calculate the 99% confidence interval for the number of chocolate chips per cookie, we will use the formula for the confidence interval when the population standard deviation is unknown, which requires the t-distribution:

\[ \text{Confidence Interval} = \bar{x} \pm t^* \left(\frac{s}{\sqrt{n}}\right) \]

Where:

  • \(\bar{x} = 6.3\) (sample mean)
  • \(s = 3.1\) (sample standard deviation)
  • \(n = 46\) (sample size)
  • \(t^*\) is the critical t-value for a 99% confidence level, with degrees of freedom \(df = n - 1 = 46 - 1 = 45\).
  1. Finding the critical t-value (\(t^*\)): To find the critical t-value for a 99% confidence level and 45 degrees of freedom, we can refer to a t-table or use statistical software/calculator. The critical t-value for 99% confidence level (two-tailed) is approximately:

    \(t^* \approx 2.680\)

  2. Calculating the standard error (SE): \[ SE = \frac{s}{\sqrt{n}} = \frac{3.1}{\sqrt{46}} \approx \frac{3.1}{6.782} \approx 0.456 \]

  3. Calculating the margin of error (ME): \[ ME = t^* \times SE = 2.680 \times 0.456 \approx 1.224 \]

  4. Calculating the confidence interval:

    • Lower limit: \[ \bar{x} - ME = 6.3 - 1.224 \approx 5.1 \]
    • Upper limit: \[ \bar{x} + ME = 6.3 + 1.224 \approx 7.5 \]

Therefore, the 99% confidence interval for the number of chocolate chips per cookie for Big Chip cookies is:

\[ (5.1, 7.5) \]

In summary, rounding to one decimal place:

The 99% confidence interval is (5.1, 7.5).