Question

in a random sample of four microwave ovens, the mean repair cost was 65.00 and the standard deviation was 12.50 assume the population is normally distributed and use a t-distribution to construct a 95% confidence interval for the population mean u what is the margin of error of u? interpret the results

In a Round to two decimal places as needed.)

Answer 1

Sample size n = 4

Degrees of freedom = n-1= 3

The critical value of t for 3 df with 95% Confidence is +/-3.182

The 95% Confidence Interval for $\mu$ is

$\overline{x}\pm t\times \frac{s}{\sqrt{n}} = 65\pm 3.182\times \frac{12.5}{\sqrt{4}}$

$= 65\pm 19.8875$

The 95% Confidence Interval for $\mu$ is ( 45.11, 84.89)

b) The Margin of Error is

$t\times \frac{s}{\sqrt{n}} = 3.182\times \frac{12.5}{\sqrt{4}} = 19.89$

c) Interpretation : Option B) If a large sample of microwave are takken approximately 95% of them will have repair costs between the bounds of the confidence interval.