Problem of statistical inference Poisson

Question

I am having problems solving this problem of statistical inference and I do not know if it is well done or not, so I would like someone to review it. I just started with inference, so I have my problems.

-The weekly number of failures due to software problems that have occurred in a computer system are the following:

Number fails 0  1  2  3  4  5  6  8
Frequency   41 62 63 38 12  7  1  1

1) Can it be accepted that the weekly number of failures due to software problems that occur in that computer system is a random variable with Poisson distribution?

What I have to do is calculate the chi-square, which I get from Chi-Square values ​​= 379,682 with 11 g.l. Value-P = 0,0 It does not fit me since it gives me a p-value of 0 and it never happened to me, so I do not know if you could tell me that this result is true

2) Assuming that the weekly number of failures due to software problems is distributed according to a law Poisson: Calculate an unbiased estimate of the weekly rate of system failures due to software problems.

What I have to calculate is a mean that the best estimator (and also unbiased is the sample mean) that has a value of 1.77333 to be unbiased E (Estimation) = μ and because E (Average) = μ We concluded that he would be unbiased.

3) Calculate a level confidence interval of at least 0.94 for the distribution parameter.

I have this, but I am not very confident that it is correct:

Average (x): 1,77333
Standard deviation (s): 1.36185
(x +/- Ks)
100 (1-1 / K ^ 2) = 94; K = 4.0824
1.7733 +/- 4.0824 * 1.36185 = [-3.7863, 7,3329]

4) Calculate a confidence interval of approximate level 0.94 for the distribution parameter using the central limit theorem.

Of the sample rate I have my own mean which is 1.7733 and of null hypothesis, as they do not tell me anything, I suppose 0.5 for n = 225 and I get the IC [1.61016, 1.94868]

5) Can it be accepted, at a level of 2%, that, as the engineer responsible for the system says, the weekly failure rate is at most 1.6? And at a level of 5%?

It is the same exercise as the previous one but with a null hypothesis of 1.6, for an average of 1.7733 and n = 225. As it tells me that it is unilateral less than (HERE I HAVE DOUBTS IF IT IS UNILATERAL LESS THAN) and with the p-value of 0.981321 I conclude that I can not reject the null hypothesis for alpha 0.02.

For 5% I get the same p-value and same conclusion of rejection

Answer 1

Here is an outline of the procedure you need to follow:

Goodness-of-fit to Poisson. Let the frequencies $f_i = 41, 62, 63, \dots$ and the values be $k_i = 0, 1, 2, \dots .$ Then estimate the population mean as $\hat \lambda = \frac 1 n \sum_{i=1}^9 k_i f_i = 1.7733,$ where $n = \sum_{i=1}^9 f_i = 225.$

Then use observed category counts $X_j = 0, 1, 2, 3, 4,$ and $\ge 5,$ where the last count is $7 = 1 + 0 + 1 = 9.$ Then according to the distribution $\mathsf{Pois}(\hat \lambda = 1.7733),$ find the probabilities of each of the categories, and multiply them by the sample size $n$ to obtain expected counts $E_j.$

Contributions $C_j = \frac {(X_j - E_j)^2}{E_j}$ to the chi-squared statistic $Q = \sum_j C_j = 2.096$ are shown in the table below along with the $X_j$ and $E_j.$

     X         E         C
    41 38.198659 0.2054395
    62 67.737681 0.4860070
    63 60.059615 0.1439547
    38 35.501239 0.1758758
    12 15.738587 0.8880740
     9  7.764219 0.1966912

Under the null hypothesis that data are Poisson, the 'chi-squared' statistic $Q$ is approximately distributed according to the distribution $\mathsf{Chisq}(\text{df} = 6-2 = 4).$ The null hypothesis is not rejected at the 5% level because the critical value for the test is $c = 9.488 > 2.096;$ you can find $c$ in printed tables of the chi-squared distribution.

It was necessary to combine original counts 7, 1, 0, 1 into one category in order to have the last of the $E_j$'s exceed $5,$ assuring a reasonable fit of the null distribution to chi-squared.

I have tried to be careful with my computations in this procedural outline, but you should check them.

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Confidence interval for Poisson mean. I am not sure what style of confidence interval you are using for the Poisson mean $\lambda$ based on the estimate $\hat \lambda;$ I don't know what you mean by 'of at least 0.94', I don't see the rationale for your value of $k,$ and I can't follow all of your computer code. Also, I'm suspicious of the validity of a CI for $\lambda > 0$ with a lower endpoint that strays so far into negative territory.

A Wald 95% CI for $n\lambda$ is of the form $n\hat\lambda \pm 1.96\sqrt{n\hat\lambda}.$ Divide the endpoints by $n$ to get a 95% CI $(1.610, 1.937)$ for $\lambda.$ [This is an asymptotic interval and so it should be reasonably accurate for a sample of size as large as $n = 255.]$

It has been suggested that a more accurate CI for $n\lambda$ is of the form $n\hat\lambda + 2 \pm 1.96\sqrt{n\hat\lambda + 1},$ but for $n = 255,$ the adjustments don't make much difference. The corresponding result for $\lambda$ is $(1.618, 1.945).$

There must be at least half a dozen styles of Poisson CIs in use, each with enthusiastic adherents, and I don't claim mine is 'best'. Perhaps we will get some recommendations from others on this site.

A simple simulation (in R) seems to show that this interval has about 95% actual coverage for $\lambda$ and $n$ of the sizes we have here.

set.seed(1201)
lam = 1.7733;  n = 255
lam.tot = replicate(10^5, sum(rpois(n, lam)))
lcl = (lam.tot + 2 - 1.96*sqrt(lam.tot + 1))/n
ucl = (lam.tot + 2 + 1.96*sqrt(lam.tot + 1))/n
mean(lcl < lam & ucl > lam)
[1] 0.94997