Let's walk through each part of the question step-by-step.
### Given Data:
- Sample values: [tex]\(9, 14, 10, 12, 7, 13, 12\)[/tex]
- Sample mean from hypothesis test: 49
- Population mean: 50
- Population standard deviation ([tex]\(\sigma\)[/tex]): 4
- Sample size ([tex]\(n\)[/tex]): 100
- Confidence level: 95%
#### Part a) Point Estimate of [tex]\(\mu\)[/tex]
The point estimate of [tex]\(\mu\)[/tex] (population mean) is simply the sample mean.
To calculate the sample mean ([tex]\(\bar{x}\)[/tex]):
[tex]\[
\bar{x} = \frac{\sum \text{values}}{n} = \frac{9 + 14 + 10 + 12 + 7 + 13 + 12}{7} = \frac{77}{7} = 11
\][/tex]
So, the point estimate of [tex]\(\mu\)[/tex] is 11.
#### Part b) 95% Confidence Interval for [tex]\(\mu\)[/tex]
To find the 95% confidence interval for [tex]\(\mu\)[/tex], we need to follow these steps:
1. Calculate the sample mean of the given values.
[tex]\[
\bar{x}_{\text{values}} = \frac{9 + 14 + 10 + 12 + 7 + 13 + 12}{7} = 11
\][/tex]
2. Calculate the sample standard deviation (s):
[tex]\[
s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}
\][/tex]
Let's calculate it step-by-step:
[tex]\[
\begin{aligned}
(x_i - \bar{x})^2 & \\
(9-11)^2 & = 4 \\
(14-11)^2 & = 9 \\
(10-11)^2 & = 1 \\
(12-11)^2 & = 1 \\
(7-11)^2 & = 16 \\
(13-11)^2 & = 4 \\
(12-11)^2 & = 1 \\
\end{aligned}
\][/tex]
[tex]\[
\sum (x_i - \bar{x})^2 = 4 + 9 + 1 + 1 + 16 + 4 + 1 = 36
\][/tex]
[tex]\[
s = \sqrt{\frac{36}{6}} = \sqrt{6} \approx 2.449
\][/tex]
3. Determine the z-score for a 95% confidence interval (1.96):
The critical value [tex]\( z \)[/tex] for a 95% confidence level is approximately 1.96.
4. Calculate the margin of error (ME):
[tex]\[
ME = z \cdot \frac{s}{\sqrt{n}} = 1.96 \cdot \frac{2.449}{\sqrt{7}} \approx 1.815
\][/tex]
5. Build the confidence interval:
[tex]\[
\bar{x} \pm ME = 11 \pm 1.815 \Rightarrow (9.185, 12.815)
\][/tex]
So, the 95% confidence interval for [tex]\(\mu\)[/tex] is approximately (9.185, 12.815).
#### Hypothesis Testing
We are testing the hypothesis:
- Null Hypothesis ([tex]\(H_0\)[/tex]): [tex]\(\mu = 50\)[/tex]
- Alternative Hypothesis ([tex]\(H_1\)[/tex]): [tex]\(\mu < 50\)[/tex]
We use the z-test formula for this:
[tex]\[
z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} = \frac{49 - 50}{\frac{4}{\sqrt{100}}} = \frac{49 - 50}{0.4} = -2.5
\][/tex]
To find the p-value associated with a z-score of -2.5, we look up the cumulative distribution function (CDF) for the normal distribution:
[tex]\[
p = \Phi(-2.5) \approx 0.0062
\][/tex]
Since the p-value (0.0062) is less than the significance level of 0.05, we reject the null hypothesis. Thus, the sample provides sufficient evidence to support the manager's claim that the average content of juice per bottle is less than 50 cl.
