Cfrgtkky

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

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.

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 $nlambda$ is of the form $nhatlambda pm 1.96sqrt{nhatlambda}.$ 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 $nlambda$ is of the form
$nhatlambda + 2 pm 1.96sqrt{nhatlambda + 1},$ 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