Question

The amounts (in ounces) of juice in eight randomly selected juice bottles are: 15.8, 15.6, 15.1, 15.2, 15.1, 15.5, 15.9, 15.5 construct a 95% confidence interval for the mean amount of juice in all such bottles. assume an approximate normal distribution.
a.(15.250, 15.675)
b.(15.206, 15.719)
c.(15.257, 15.668)
d.(15.284, 15.641)
e.none of the above

Answer 1

The confidence interval is given by the formula:
m +/- z·(σ/√n)

First, compute the mean:
m = (15.8 + 15.6 + 15.1 + 15.2 + 15.1 + 15.5 + 15.9 + 15.5) / 8
    = 15.463

Then, compute the standard deviation:
σ = √[∑(v - m)²/n]
   = 0.287

The z-score for a 95% confidence interval is z = 1.96.

Now you can calculate:
m + z·(σ/√n) = 15.463 + 1.96·(0.287/√8)
                        = 15.463 + 0.199
                        = 15.662

m - z·(σ/√n) = 15.463 - 1.96·(0.287/√8)
                        = 15.463 - 0.199
                        = 15.264

Therefore the confidence interval is (15.264, 15.662) and the correct answer is E) none of the above.

Answer 2

The $95\%$ confidence interval for the population mean is $\left( {15.250,15.675}\right).$

Further explanation:

The formula for confidence interval can be expressed as follows,

$\boxed{{\text{Confidence interval}} = \left( {\overline X\pm ME} \right)}$

ME represents the Margin of Error.

The formula of margin of error is $\boxed{{\text{ME}} = {Z_{\dfrac{\alpha }{2}}}\times\dfrac{\sigma}{{\sqrt n }}}.$

Here, Z is the standard normal value, $\sigma$ is the standard deviation and n represents the total observation.

Explanation:

The data values are $15.8, 15.6, 15.1, 15.2, 15.1, 15.5, 15.9$ and $15.5.$

The mean can be calculated as follows,

$\begin{aligned}{\text{Mean}}\left({\overline X } \right)&= \frac{{15.8 + 15.6 + 15.1 + 15.2 + 15.1 + 15.5+ 15.9 + 15.5}}{8}\\&= \frac{{123.7}}{8}\\&= 15.46\\\end{aligned}$

The formula of standard deviation can be expressed as,

$\boxed{s = \sqrt{\frac{{\sum {{{\left( {{x_i}-\overline x }\right)}^2}} }}{{n - 1}}}}$

The standard deviation is $0.307.$

The margin of error can be obtained as follows.

$\begin{aligned}{\text{Confidence interval}}&= 15.46 \pm 1.96 \times \frac{{0.307}}{{\sqrt8 }}\\&= 15.46 \pm 0.210\\ &= \left( {15.250,15.675} \right)\\\end{aligned}$

The $95\%$ confidence interval for the population mean is $\left( {15.250,15.675}\right).$

Learn more:

1. Learn more about normal distribution brainly.com/question/12698949

2. Learn more about standard normal distribution brainly.com/question/13006989

3. Learn more about confidence interval of meanhttps://brainly.com/question/12986589

Answer details:

Grade: College

Subject: Statistics

Chapter: Confidence Interval

Keywords:  Z-score, eight, amount, ounce, juice bottle, confidence interval, confidence limit, 95 percent, standard normal distribution, standard deviation, test, measure, probability, low score, mean, repeating, indicated, normal distribution, percentile, percentage, undesirable behavior, proportion, empirical rule.