A manufacturer is interested in estimating the lifetime of printers. A confidence interval is constructed based on a sample of 20 printers and the assumption that the population distribution of the lifetime of a printer is normal.
1. UL = x + z*(σ/√n)
6.36 = x + 1.96*(1.5/√20)
x = 6.36 - 1.96*(1.5/√20) = 5.703
2. LL = x - t*(s/√n)
3.14 = x - 1.73*(1.2/√20)
x = 3.14 + 1.73*(1.2/√20) = 3.604
3. The hypothesis being tested is:
Null hypothesis | H₀: σ² = 1.8 |
Alternative hypothesis | H₁ : σ² ≠ 1.8 |
Method | Test Statistic |
DF | P-Value |
Chi-Square | 15.20 | 19 | 0.580 |
The p-value is 0.580.
Since the p-value (0.580) is greater than the significance level (0.10), we fail to reject the null hypothesis.
Therefore, we can conclude that ?2 = 1.8.
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