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A manufacturer of sprinkler systems used for re protection in once buildings claims that the true 130F. average system-activation temperature is A sample of 9 systems when tested, yields a sample 131:08F. average activation temperature of Assume that the distribution of activation times is normal 1:5F. with a population standard deviation of a) Does the data contradict the claim at 0.01 level? Carry out a detailed test by writing the appropriate hypotheses, rejection region and conclusion.

5. A manufacturer of sprinkler systems used for fire protection in office buildings claims that the true average system-activ b) Find a 95% CI for then true average system-activation temperature.

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Answer #1

A] Ho: μ=130

Ha: μ≠130

This corresponds to a two-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

Rejection Region: Based on the information provided, the significance level is α=0.01, and the critical value for a two-tailed test is zc=2.58.

The rejection region for this two-tailed test is R={z:∣z∣>2.58}

(3) Test Statistics

The z-statistic is computed as follows:

X - μo 2 = 131.08 – 130 1.5/9 2.16 σ/νη

Decision about the null hypothesis

Since it is observed that ∣z∣=2.16≤zc​=2.58, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.0308, and since p=0.0308≥0.01, it is concluded that the null hypothesis is not rejected.

Conclusion: It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μ is different than 130, at the 0.01 significance level.

B] Confidence Interval

The 99% confidence interval is 129.792<μ<132.368.

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