Question

# In a random sample of six mobile​ devices, the mean repair cost was \$75.00 and the...

In a random sample of six mobile​ devices, the mean repair cost was \$75.00 and the standard deviation was \$13.0013.00. Assume the population is normally distributed and use a​ t-distribution to find the margin of error and construct a 95​% confidence interval for the population mean. Interpret the results.

The 95​% confidence interval for the population mean μ is

Solution :

Given that,

Point estimate = sample mean = = \$75.00

sample standard deviation = s = \$13.00

sample size = n = 6

Degrees of freedom = df = n - 1 = 6 - 1 = 5

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t /2,df = t0.025,5 = 2.571

Margin of error = E = t/2,df * (s /n)

= 2.571 * (13.00 / 6)

Margin of error = E = 13.64

The 95% confidence interval estimate of the population mean is,

- E < < + E

75.00 - 13.64 < < 75.00 + 13.64

61.36 < < 88.64

The 95​% confidence interval for the population mean μ is (61.36 , 88.64)

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