Question

One factory claims that its remote control carts last an average of 320 hours.
To verify this average, a sample of 15 carts is taken. If the calculated value falls within a symmetrical range from the 90% average, the factory is satisfied.
What conclusion would the company obtain according to this sample? The sample
were the following data:

320, 321, 311, 313, 310, 300, 302, 312, 310, 275, 310, 287, 293, 300, 295

Answer 1

The computations are done in the following table here:

X	(X - Mean(X))^2
320	258.14
321	291.27
311	49.94
313	82.20
310	36.80
300	15.47
302	3.74
312	65.07
310	36.80
275	837.14
310	36.80
287	286.74
293	119.54
300	15.47
295	79.80

The sample mean here is computed as:

$\bar X = \frac{\sum X_i}{n} = \frac{320 + 321 + ... + 295}{15} = 303.93$

The sample standard deviation is now computed here as: (from second column)

$s = \sqrt{\frac{\sum (X_i - \bar X)^2}{n-1}} = \sqrt{\frac{2214.93}{15-1}} = 12.5781$

For 90% confidence level, and n - 1 = 14 degrees of freedom, we have from t distribution tables here:
P( t₁₄ < 1.761) = 0.95

therefore, due to symmetry, we have here:
P( - 1.761 < t₁₄ < 1.761) = 0.9

Therefore the confidence interval here is computed as:

S FX n

12.5781 303.93±1.761 水 15

$303.93 \pm 5.72$

As the given value of 320 does not lie in the above computed confidence interval, therefore the factory is not satisfied here.