Question

the number of fruit snacks in each package is assumed to have 13 pieces per package, on average. danna opened up 32 packs of fruit snacks and found the average number of snacks to be 11.5 with a standard deviation of 1.13.

(df =)

round answers to three decimal places.

(se =)

(t =)

what is the probability of finding a sample of size 32 with an average of 11.5 fruit snacks or less?

enter answer as a decimal to 4 decimal places

use the (t)-distribution tool to answer the question and enter the (t) score for (x).

Explanation:

Identify given parameters

We extract the given values from the problem description:

• Hypothesized population mean \(\mu = 13\)
• Sample size \(n = 32\)
• Sample mean \(\bar{x} = 11.5\)
• Sample standard deviation \(s = 1.13\)

Calculate degrees of freedom

Using the t-Distribution concept:

$$ df = n - 1 = 32 - 1 = 31 $$

Calculate standard error

Using the Standard Error concept:

$$ SE = \frac{s}{\sqrt{n}} = \frac{1.13}{\sqrt{32}} \approx \frac{1.13}{5.656854} \approx 0.199758 $$

Rounding to three decimal places:

$$ SE \approx 0.200 $$

Calculate t-statistic

Using the One-Sample t-Test concept:

$$ t = \frac{\bar{x} - \mu}{SE} = \frac{11.5 - 13}{0.199758} = \frac{-1.5}{0.199758} \approx -7.50913 $$

Rounding to three decimal places:

$$ t \approx -7.509 $$

Find the cumulative probability

We find the probability \(P(T \le -7.509)\) for a \(t\)-distribution with \(df = 31\).
Since the \(t\)-score is extremely large in magnitude and negative, the probability is extremely close to 0.
Using standard statistical tables or software for \(t(31)\):

$$ P(T \le -7.509) \approx 0.0000 $$

Rounding to four decimal places:

$$ P \approx 0.0000 $$

Answer:

Question 1

The degrees of freedom is <blank>31</blank>.

Question 2

The standard error is <blank>0.200</blank>.

Question 3

The \(t\)-score is <blank>-7.509</blank>.

Question 4

The probability of finding a sample of size 32 with an average of 11.5 fruit snacks or less is <blank>0.0000</blank>.