Question

In: Statistics and Probability

11_ The average weight of 40 randomly selected minivans was 4150 pounds. The standard deviation was...

11_ The average weight of 40 randomly selected minivans was 4150 pounds. The standard deviation was 480 pounds. Find the 99% confidence interval of the true mean weight of the minivans.

What is the standard error for step1? What is the Z or t score for step 2? What is the confidence interval (Step 3)

Expert Solution

Point estimate = sample mean = = 4150

sample standard deviation = s = 480

sample size = n = 40

Degrees of freedom = df = n - 1 = 39

1)

standard error

_S = s / n = 480 / 40 = 75.8947

2)

At 99% confidence level the t is ,

t_/2,df = t_0.005,39 = 2.708

Margin of error = E = t_/2,df * (s / $\sqrt$ n)

= 2.708* (480 / $\sqrt$ 40)

= 205.523

3)

The 99% confidence interval estimate of the population mean is,

- E < < + E

4150 - 205.523 < < 4150 + 205.523

3944.477 < < 4355.523

(3944.477 , 4355.523)

orchestra answered 1 year ago

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