Respuesta :
Answer:
(0.9464,1.034)
Step-by-step explanation:
First find the mean μ of all values as;
(0.95+1.02+ 1.01+ 0.98 ) /4 =0.99
μ=0.99
Find the standard deviation as;
-For each value subtract the mean and square the results
0.95-0.99 = -0.04 , -0.04² = 0.0016
1.02-0.99= 0.03, 0.03²=0.0009
1.01-0.99= 0.02, 0.02²= 0.0004
0.98-0.99 = -0.01, -0.01²=0.0001
Find mean of squared deviations as;
(0.0016+0.0009+0.0004+0.0001)/4 =0.00075 ----variance
Standard deviation = √0.00075 = 0.0274
δ = 0.0274
n=4
Thus n< 30 , confidence interval for μ = x ± t*δ/√n ---------(i)
Degree of freedom (df) = n-1 =4-1 =3
The t value for 95% confidence with df= 3 is t=3.182
Substituting the values in equation (i)
0.99 ± 3.182 * 0.0274/√4
0.99 ± 3.182 * 0.0274/2
0.99 ± 3.182 * 0.0137
0.99 ± 0.0436
(0.9464,1.034)
[tex]95[/tex]% confidence interval for [tex]\mu \ is \ (0.9397,1.0403)[/tex].
To understand the calculations, check below.
Confidence interval:
The confidence level represents the proportion (frequency) of acceptable confidence intervals that contain the true value of the unknown parameter. So that the proportion of the range contains the true value of the parameter that will be equal to the confidence level.
Measurements in grams: [tex]x_{i} =0.95,1.02,1.01,0.98[/tex]
Mean of measurements: Sample mean:[tex]\bar{x}=\frac{\sum x}{n}=\frac{0.95+1.02+1.01+0.98}{4}[/tex]
[tex]=\frac{3.96}{4}=0.99[/tex]
Sample size: [tex]4[/tex]
The standard deviation of measurements: Sample Standard deviation: [tex]S=\sqrt{\frac{\sum \left ( x-\bar{x} \right )^{2}}{n-1}} \\ =\sqrt{\frac{{\left ( 0.95-0.99 \right )^{2}+\left ( 1.02-0.99 \right)^{2}+\left ( 1.01-0.99 \right )^{2}+\left ( 0.98-0.99 \right )^{2}} }{4-1}} \\ =\sqrt{0.001}=0.0316[/tex]
Confidence interval for [tex]\mu[/tex]
[tex]\bar{x}\pm t_{a/2}\times \frac{s}{\sqrt{n}}[/tex]
[tex]\alpha =\frac{\left ( 100-95 \right )}{100}=0.05;\frac{\alpha }{2}=0.025[/tex]
Degrees of freedom = sample size[tex]-1=n-1=4-1=3[/tex]
[tex]t_{0.025}[/tex] for [tex]3[/tex] degrees of freedom[tex]=3.182[/tex]
[tex]95[/tex]% confidence interval for [tex]\mu[/tex]
[tex]\bar{x}\pm t_{0.025} \times\frac{s}{\sqrt{n} } =0.99\pm3.182\times\frac{0.0316}{\sqrt{4} \\}\\=0.99\pm0.0503\\=(0.9397,1.0403)[/tex]
Learn more about the topic of Confidence interval: https://brainly.com/question/23630128
