Respuesta :
Answer:
[tex]69.72-1.71\frac{4.15}{\sqrt{25}}=68.30[/tex]
[tex]69.72+1.71\frac{4.15}{\sqrt{25}}=71.14[/tex]
Step-by-step explanation:
Information given
[tex]\bar X=69.72[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
s=4.15 represent the sample standard deviation
n=25 represent the sample size
Confidence interval
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
The degrees of freedom, given by:
[tex]df=n-1=25-1=24[/tex]
Since the Confidence is 0.90 or 90%, the significance is [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and the critical value is [tex]t_{\alpha/2}=1.71[/tex]
Now we have everything in order to replace into formula (1):
[tex]69.72-1.71\frac{4.15}{\sqrt{25}}=68.30[/tex]
[tex]69.72+1.71\frac{4.15}{\sqrt{25}}=71.14[/tex]
Using the t-distribution, it is found that the 90% confidence interval for the mean height of adult males in the United States is (68.3, 71.14).
We are given the standard deviation for the sample, which is why the t-distribution is used to solve this question.
The information given is:
- Sample mean of [tex]\overline{x} = 69.72[/tex].
- Sample standard deviation of [tex]s = 4.15[/tex].
- Sample size of [tex]n = 25[/tex].
The confidence interval is:
[tex]\overline{x} \pm t\frac{s}{\sqrt{n}}[/tex]
The critical value, using a t-distribution calculator, for a two-tailed 90% confidence interval, with 25 - 1 = 24 df, is t = 1.7109.
Then:
[tex]\overline{x} - t\frac{s}{\sqrt{n}} = 69.72 - 1.7109\frac{4.15}{\sqrt{25}} = 68.3[/tex]
[tex]\overline{x} + t\frac{s}{\sqrt{n}} = 69.72 + 1.7109\frac{4.15}{\sqrt{25}} = 71.14[/tex]
The 90% confidence interval for the mean height of adult males in the United States is (68.3, 71.14).
A similar problem is given at https://brainly.com/question/15180581
