Respuesta :
Answer:
[tex]3.3\leq x\leq 3.5[/tex]
Step-by-step explanation:
The 95% confidence interval can be calculated as:
[tex]x'-z_{\alpha/2}\frac{s}{\sqrt{n}} \leq x\leq x'+z_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]
Where [tex]x[/tex] is the mean consumption of milk, [tex]x'[/tex] is the sample mean, [tex]s[/tex] is the population standard deviation, [tex]n[/tex] is the size of the sample, [tex]\alpha[/tex] is 5% and [tex]z_{\alpha /2}[/tex] is the z-value that let a probability of [tex]\alpha /2[/tex] on the right tail.
So, replacing [tex]x'[/tex] by 3.4 liters, [tex]s[/tex] by 1.3 liters, n by 650 and [tex]z_{\alpha /2}[/tex] by 1.96, we get that the 95% confidence interval is equal to:
[tex]3.4-(1.96*\frac{1.3}{\sqrt{650}})\leq x\leq 3.4+(1.96*\frac{1.3}{\sqrt{650}}) \\3.4-0.1\leq x\leq 3.4+0.1\\3.3\leq x\leq 3.5[/tex]
Answer:
Step-by-step explanation:
Confidence interval is written in the form,
(Sample mean - margin of error, sample mean + margin of error)
The sample mean, x is the point estimate for the population mean.
Since the sample size is large and the population standard deviation is known, we would use the following formula and determine the z score from the normal distribution table.
Margin of error = z × σ/√n
Where
σ = population standard Deviation
n = number of samples
From the information given
x = 3.4
σ = 1.3
n = 650
To determine the z score, we subtract the confidence level from 100% to get α
α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025
This is the area in each tail. Since we want the area in the middle, it becomes
1 - 0.025 = 0.975
The z score corresponding to the area on the z table is 1.96. Thus, confidence level of 95% is 1.96
Margin of error = 1.96 × 1.3/√650 = 0.1
Confidence interval = 3.4 ± 0.1
The lower end of the confidence interval is
3.4 - 0.1 = 3.3 litres
The upper end of the confidence interval is
3.4 + 0.1 = 3.5 litres
