Question

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the number of fruit snacks in each package is assumed to have 13 pieces per package, on average. destiny opened up 20 packs of fruit snacks and found the average number of snacks to be 11.7 with a standard deviation of 1.19.

\\(df =\\)

round answers to three decimal places.

\\(se =\\)

\\(t =\\)

what is the probability of finding a sample of size 20 with an average of 11.7 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 the given parameters

We extract the sample statistics and population parameter from the problem statement:

• Null hypothesis mean \(\mu_0 = 13\)
• Sample size \(n = 20\)
• Sample mean \(\bar{x} = 11.7\)
• Sample standard deviation \(s = 1.19\)

Calculate degrees of freedom

The degrees of freedom \(df\) for a one-sample \(t\)-test is:

$$ df = n - 1 = 20 - 1 = 19 $$

Calculate the standard error

The standard error \(SE\) of the sample mean is:

$$ SE = \frac{s}{\sqrt{n}} = \frac{1.19}{\sqrt{20}} $$

$$ SE \approx \frac{1.19}{4.472136} \approx 0.266092 $$

Rounding to three decimal places:

$$ SE \approx 0.266 $$

Calculate the t-statistic

The \(t\)-score is calculated using:

$$ t = \frac{\bar{x} - \mu_0}{SE} = \frac{11.7 - 13}{0.266092} $$

$$ t = \frac{-1.3}{0.266092} \approx -4.885528 $$

Rounding to three decimal places:

$$ t \approx -4.886 $$

Find the cumulative probability

We find the probability \(P(T \le t)\) for \(df = 19\) and \(t = -4.886\):
Using a standard \(t\)-distribution table or calculator:

$$ P(T \le -4.886) \approx 0.000052 $$

Rounding to four decimal places:

$$ P \approx 0.0001 $$

Answer:

\(df =\) <blank>19</blank>

\(SE =\) <blank>0.266</blank>

\(t =\) <blank>-4.886</blank>

What is the probability of finding a sample of size 20 with an average of 11.7 fruit snacks or less?
<blank>0.0001</blank>