Respuesta :
The 90% confidence interval for the population mean of the considered population from the given sample data is given by: Option C: [130.10, 143.90]
How to find the confidence interval for population mean from large samples (sample size > 30)?
Suppose that we have:
- Sample size n > 30
- Sample mean = [tex]\overline{x}[/tex]
- Sample standard deviation = s
- Population standard deviation = [tex]\sigma[/tex]
- Level of significance = [tex]\alpha[/tex]
Then the confidence interval is obtained as
- Case 1: Population standard deviation is known
[tex]\overline{x} \pm Z_{\alpha /2}\dfrac{\sigma}{\sqrt{n}}[/tex]
- Case 2: Population standard deviation is unknown.
[tex]\overline{x} \pm Z_{\alpha /2}\dfrac{s}{\sqrt{n}}[/tex]
For this case, we're given that:
- Sample size n = 90 > 30
- Sample mean = [tex]\overline{x}[/tex] = 138
- Sample standard deviation = s = 34
- Level of significance = [tex]\alpha[/tex] = 100% - confidence = 100% - 90% = 10% = 0.1 (converted percent to decimal).
At this level of significance, the critical value of Z is: [tex]Z_{0.1/2}[/tex] = ±1.645
Thus, we get:
[tex]CI = \overline{x} \pm Z_{\alpha /2}\dfrac{s}{\sqrt{n}}\\CI = 138 \pm 1.645\times \dfrac{34}{\sqrt{90}}\\\\CI \approx 138 \pm 5.896\\CI \approx [138 - 5.896, 138 + 5.896]\\CI \approx [132.104, 143.896] \approx [130.10, 143.90][/tex]
Thus, the 90% confidence interval for the population mean of the considered population from the given sample data is given by: Option C: [130.10, 143.90]
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