Question

Aisha wants to advertise how many chocolate chips are in each Big Chip cookie at her bakery. She randomly selects a sample of 76 cookies and finds that the number of chocolate chips per cookie in the sample has a mean of 8.9 and a standard deviation of 2.3. What is the 90% confidence interval for the number of chocolate chips per cookie for Big Chip cookies? Assume the data is from a normally distributed population. Round answers to 3 decimal places where possible.

Answer 1

Using the t-distribution, the 90% confidence interval for the number of chocolate chips per cookie for Big Chip cookies is (8.374, 9.426).

What is a t-distribution confidence interval?

The confidence interval is:

$\overline{x} \pm t\frac{s}{\sqrt{n}}$

In which:

• $\overline{x}$ is the sample mean.

• t is the critical value.

• n is the sample size.

• s is the standard deviation for the sample.

The critical value, using a t-distribution calculator, for a two-tailed 90% confidence interval, with 76 - 1 = 75 df, is t = 1.9921.

The other parameters are given as follows:

$\overline{x} = 8.9, s = 2.3, n = 76$.

Hence the bounds of the interval are:

$\overline{x} - t\frac{s}{\sqrt{n}} = 8.9 - 1.9921\frac{2.3}{\sqrt{76}} = 8.374$

$\overline{x} + t\frac{s}{\sqrt{n}} = 8.9 + 1.9921\frac{2.3}{\sqrt{76}} = 9.426$

More can be learned about the t-distribution at brainly.com/question/16162795

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