Respuesta :
Answer:
The 95% confidence interval to estimate is 475.4592 to 484.5408
Step-by-step explanation:
Consider the provided information.
25 pieces of this metallic glass were randomly sampled from a recent production run.
That means the value of n is 25.
The degree of freedom is:
df = n-1
df = 25-1 = 24
X = 480°F and s = 11°F
We need to Construct a 95% confidence interval to estimate.
1-α=0.95
α=0.05
It is a two tail test with small sample size.
Determine the value of t by using Degrees of freedom and Significance level:
The required t value is 2.064
[tex]95\% CI=\bar x\pm t_c\times \frac{s}{\sqrt{n}}[/tex]
Substitute the respective values as shown:
[tex]95\% CI=480\pm 2.064\times \frac{11}{\sqrt{25}}[/tex]
[tex]95\% CI=480\pm 4.5408[/tex]
[tex]95\% CI=475.4592\ to\ 484.5408[/tex]
Hence, the 95% confidence interval to estimate is 475.4592 to 484.5408
Using the t-distribution, as we have the standard deviation for the sample, it is found that the 95% confidence interval to estimate the temperature at which brittleness first appeared is given by (475.46 ºF, 484.54 ºF).
What is a t-distribution confidence interval?
The confidence interval is:
[tex]\overline{x} \pm t\frac{s}{\sqrt{n}}[/tex]
In which:
- [tex]\overline{x}[/tex] is the sample mean.
- t is the critical value.
- n is the sample size.
- s is the standard deviation for the sample.
The critical value, using a t-distribution calculator, for a two-tailed 95% confidence interval, with 25 - 1 = 24 df, is t = 2.0639.
The other parameters are as follows:
[tex]n = 25, \overline{x} = 480, s = 11[/tex]
Hence:
[tex]\overline{x} - t\frac{s}{\sqrt{n}} = 480 - 2.0639\frac{11}{\sqrt{25}} = 475.46[/tex]
[tex]\overline{x} + t\frac{s}{\sqrt{n}} = 480 + 2.0639\frac{11}{\sqrt{25}} = 484.54[/tex]
The 95% confidence interval is (475.46 ºF, 484.54 ºF).
More can be learned about the t-distribution at https://brainly.com/question/16162795
