Answer :
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.
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.
The [tex]95\%[/tex] confidence interval for the population mean is [tex]\left( {15.250,15.675}\right).[/tex]
Further explanation:
The formula for confidence interval can be expressed as follows,
[tex]\boxed{{\text{Confidence interval}} = \left( {\overline X\pm ME} \right)}[/tex]
ME represents the Margin of Error.
The formula of margin of error is [tex]\boxed{{\text{ME}} = {Z_{\dfrac{\alpha }{2}}}\times\dfrac{\sigma}{{\sqrt n }}}.[/tex]
Here, Z is the standard normal value, [tex]\sigma[/tex] is the standard deviation and n represents the total observation.
Explanation:
The data values are [tex]15.8, 15.6, 15.1, 15.2, 15.1, 15.5, 15.9[/tex] and [tex]15.5.[/tex]
The mean can be calculated as follows,
[tex]\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}[/tex]
The formula of standard deviation can be expressed as,
[tex]\boxed{s = \sqrt{\frac{{\sum {{{\left( {{x_i}-\overline x }\right)}^2}} }}{{n - 1}}}}[/tex]
The standard deviation is [tex]0.307.[/tex]
The margin of error can be obtained as follows.
[tex]\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}[/tex]
The [tex]95\%[/tex] confidence interval for the population mean is [tex]\left( {15.250,15.675}\right).[/tex]
Learn more:
1. Learn more about normal distribution https://brainly.com/question/12698949
2. Learn more about standard normal distribution https://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.