Question

A random sample of 10 chocolate bars has, on average, 230 calories/bar, with a standard deviation of 15 calories. Assume that the distribution of the calorie content population is approximately normal.

Answer 1

Answer:

We are not told what to calculate, but the following are possible:

1. Margin of error (E) = 9.30 (approx.)

2. Confidence interval (CI) = [220.70 < μ < 239.30] - Approx.

Step-by-step explanation:

We are not told what to calculate, but the following are possible:

1. Margin of error (E), and

2. Confidence interval (CI)

Given that:

Sample size (n) = 10

Mean (x bar) = 230 and,

Standard deviation (sd) = 15.

We assumed, 95% confidence level.

And is approximately normal. This implies that, we should use Z statistics instead of t statistics.

1) E = ±Z * $\frac{\sigma}{\sqrt{n} }$

  E =  ±1.96 * 4.743416 = 9.297095

  E = 9.30 (approx.)

2) CI = xbar ± E

   CI = 230  ± 9.297095

   CI = 220.7029 < μ < 239.2971

   CI = [220.70 < μ < 239.30] - Approx.