Based on past records, the main line printer in a university computer center ope

Question

Based on past records, the main line printer in a university computer center operates properly on 95% of all days. A random sample of 15 days indicates that the printer was not operation properly on three of the observed days.

A) Define the random variable and state its distribution.

B) Assuming there is no change in the proportion of days with proper operation, what is the probability of obtaining a sample with at least this many days of improper operation?

C) Based on your answer to part b, discuss whether the sample provides sufficient evidence to indicate that the proportion of days with proper operation has decreased.

Explanation / Answer

A) The random variable is samle of 15 days, and it probability distribution follws a binomial distribution.

B) The probability of obtaining a sample with at least this many days of improper operation is 0.0362.

x = 3, n = 15, p = 0.05

By applying binomial distribution:-

P(x, n, p) = nCx*px *(1 - p)(n - x)

P(x > 3) = 0.0362

C)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P > 0.05

Alternative hypothesis: P < 0.05

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected only if the sample proportion is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.

Analyze sample data. Using sample data, we calculate the standard deviation () and compute the z-score test statistic (z).

= sqrt[ P * ( 1 - P ) / n ]

= 0.0563

z = (p - P) /

z = - 2.664

where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.

Since we have a one-tailed test, the P-value is the probability that the z-score is less than - 2.66. We use the Normal Distribution Calculator to find P(z < - 2.66) = 0.0039

Interpret results. Since the P-value (0.0039) is less than the significance level (0.05), we cannot accept the null hypothesis.

From the above test we have sufficient evidence that that the proportion of days with proper operation has decreased.

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