Question

A random sample is taken and the sample size is 25 The sample is normally distributed, then sample mean is 89 , and the standard deviation is 5.5. Find a 90% confidence interval for the population mean.​

Answer 1

Answer: Lower and upper bound of confidence interval are 87.1905 and 90.8095 respectively.

Step-by-step explanation:

$CI = \mu \pm Z \frac {\sigma}{\sqrt{n}}$

CI ⇒ Confidence interval

μ  ⇒ Mean of samples

σ ⇒  Standard deviation of samples

n ⇒ Sample size

Z ⇒ Critical value of the z-distribution

For given data,

μ = 89

σ  = 5.5

n = 25

Value of Z for 90% confidence interval is 1.645

∴ $CI = \ 89 \pm 1.645 *\frac {5.5}{\sqrt{25}}$

$CI = \ 89 \pm 1.645 *\frac {5.5}{5}$

CI = 89 ± 1.8095

Lower bound of confidence interval is  87.1905

Upper bound of confidence interval is  90.8095

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