Question

Use the confidence level and sample data to find a confidence interval for estimating the population . Round your answer to the same number of decimal places as the sample mean. Test scores: n = 104, 7 = 95.3, 0 = 6.5; 99% confidence 93.7 < < 96.9 O94.1 << 96.5 93.8 < p < 96.8 94.2 << 96.4

Answer 1

Solution :

The 99% confidence interval for population mean is given as follows :

$\large \left ( \bar{x} \pm Z_{\frac{0.01}{2}}\frac{\sigma}{\sqrt{n}} \right )$

Where, x̄ is sample mean, σ is population standard deviation, n is sample size and Z_(0.01/2) is critical z-value to construct 99% confidence interval.

We have, x̄ = 95.3, σ = 6.5, n = 104

Using Z-table we get, Z_(0.01/2) = 2.576

Hence, the 99% confidence interval for population mean μ is,

$\large \left (95.3 \pm 2.576\times\frac{6.5}{\sqrt{104}} \right )$

= (95.3±1.6

$\large =\left (95.3 - 1.6, 95.3 + 1.6 \right )$

$\large =\left (93.7, 96.9 \right )$

The 99% confidence interval for population mean μ is,

93.7 < μ < 96.9

Option (1) is correct.

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