Answer to Question #260277 in Statistics and Probability for yyy

Question

A computer system analyst claims that the average number of fatal system errors that

occurred is at most 52.3 cases per day. The general manager wants to check if this

claim is true. A random sample of 100 computers taken by the general manager showed

that these computers have an average of 48.9 fatal system errors that occurred per day

with a variance of 64.

a) What is the point estimate of the population mean?

b) Find a 92% confidence interval for the corresponding population mean.

Leave your answer in 1 decimal place.

c) What is the conclusion that can be made by the general manager on the system

analyst’s claim at 2.5% significance level?

Answer 1

a) A point estimate of the mean of a population is determined by calculating the mean of a sample drawn from the population

$$\bar{x}=48.9$$

b) The critical value for $\alpha = 0.08$ and $df = n-1 = 99$ degrees of freedom is$t_c = z_{1-\alpha/2; n-1} = 1.768842.$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-t_c\times\dfrac{s}{\sqrt{n}}, \bar{x}+t_c\times\dfrac{s}{\sqrt{n}})$$

$$=(48.9-1.768842\times\dfrac{8}{\sqrt{100}}, 48.9+1.768842\times\dfrac{8}{\sqrt{100}})$$

$$=(47.5, 50.3)$$

Therefore, based on the data provided, the 92% confidence interval for the population mean is $47.5 < \mu < 50.3,$ which indicates that we are 92% confident that the true population mean $\mu$

is contained by the interval $(47.5, 50.3).$

c) The following null and alternative hypotheses need to be tested:

$H_0: \mu\leq 52.3$

$H_1: \mu>52.3$

This corresponds to a right-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

Based on the information provided, the significance level is $\alpha = 0.025,df=99$ degrees of freedom, and the critical value for a right-tailed test is $t_c =1.984217.$

The rejection region for this right-tailed test is $R = \{t: t > 1.984217\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{48.9-52.3}{8/\sqrt{100}}=-4.25$$

Since it is observed that $t = -4.25 \le1.984= t_c = 1.984,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value for right-tailed, $\alpha=0.025, df=99, t=-4.25$ is $p=0.999976,$ and since $p=0.999976>0.025=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$ is greater than 52.3, at the  $\alpha = 0.025$ significance level.

Therefore, there is enough evidence to claim that the average number of fatal system errors that

occurred is at most 52.3 cases per day, at the  $\alpha = 0.025$ significance level.