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Installation of a certain hardware takes random time with a standard deviation of 5 minutes.

(a) A computer technician installs this hardware on 64 different computers with an average installation time of 42 minutes. Compute a 95% confidence interval for the mean installation time.

(b) Suppose that the population mean installation time is 40 minutes. A technician installs the hardware on your PC. What is the probability that the installation time will be within the interval computed in (a)?

(c) A manager questions the assumption of the above. Her pilot sample of 40 installation times has a sample standard deviation of s = 6.2 min, and she says that it is significantly different from the assumed value of standard deviation = 5 min. Do you agree with the manager? Conduct a suitable test of a standard deviation.

User Jaleela
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Final answer:

The 95% confidence interval for the mean installation time is (40.775, 43.225) minutes. If the population mean is 40 minutes, there's a high probability that the installation time will fall within this range. To confirm the manager's claim about the standard deviation, a chi-square test must be conducted.

Step-by-step explanation:

Calculation of Confidence Interval and Probability

(a) To calculate the 95% confidence interval for the mean installation time with a sample mean of 42 minutes, a standard deviation of 5 minutes, and sample size of 64, we use the formula for the confidence interval of the mean with known standard deviation (Z-distribution):

CI = ± Z * (σ / √ n), where σ is the standard deviation, n is the sample size, and Z is the Z-value corresponding to the 95% confidence level (Z = 1.96).

Margin of Error (E) = 1.96 * (5 / √ 64) = 1.96 * 0.625 = 1.225.

The confidence interval is 42 ± 1.225, so (40.775, 43.225).

(b) If the population mean installation time is 40 minutes, the probability that an installation time falls within the computed interval is determined by the standard normal distribution since our sample size is large enough. Because 40 minutes is within our interval, it has a high probability of occurrence, assuming a normal distribution.

(c) To check if the new sample standard deviation (s = 6.2 min) is significantly different from 5 min, we would conduct a chi-square test for standard deviation. The null hypothesis (H0) would be that the sample variance is equal to the square of the assumed standard deviation (s^2 = 5^2). Without the actual calculations, if the chi-square value is greater than the critical value at the chosen significance level, we reject H0, supporting the manager's claim.

User Edz
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