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An engineer measured the diameter of bearings in cm. 10.2 16.9 13.3 12.5 12.1 10.0 10.9 13.5 16.6 11.1 12.0 16.2 12.3 11.8 13
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Answer #1

x = 12.83

s = 2.19

(a) t-critical value = 1.76

= x ± t*(s/√n)

= 12.83 ± 1.76*(2.19/√15)

= 11.8307, 13.8226

(b) We are 90% confident that the population mean diameter of bearings is between 11.8307 and 13.8226.

(c) The hypothesis being tested is:

H0: µ = 14

Ha: µ ≠ 14

The test statistic, t = (x - µ)/s/√n

t = (12.83 - 14)/2.19/√15 = -2.075

The critical value is 2.145.

Since 2.075 > 2.145, we fail to reject the null hypothesis.

Therefore, we cannot conclude that µ ≠ 14.

(d) The hypothesis being tested is:

H0: µ = 14

Ha: µ < 14

The test statistic, t = (x - µ)/s/√n

t = (12.83 - 14)/2.19/√15 = -2.075

The p-value is 0.0285.

Since the p-value (0.0285) is less than the significance level (0.10), we can reject the null hypothesis.

Therefore, we can conclude that µ < 14.

(e) Since 14 is above the confidence interval's upper limit, we can say that µ < 14.

(f) We cannot conclude that µ = 14.

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