Answer: (a) \(m=69.522\), \(n=0.157\) to 3 decimal places.

(b) The test statistic is \(\chi^2\approx 7.16\). With 3 degrees of freedom, the 5% critical value is \(7.815\). Since \(7.16\lt 7.815\), there is insufficient evidence to reject the expert's claim. The Poisson distribution with mean \(0.7\) is a satisfactory model for these data at the 5% significance level.

(a) Let \(X\) be the number of breakdowns in a day. Under the expert's claim, \(X\sim \operatorname{Po}(0.7)\), so \(P(X=r)=e^{-0.7}\dfrac{0.7^r}{r!}\).

The expected frequency for 1 breakdown is

\(m=200P(X=1)=200e^{-0.7}(0.7)=69.522\) to 3 decimal places.

For the final column,

\(n=200P(X\geq 5)=200\left(1-\sum_{r=0}^{4}e^{-0.7}\dfrac{0.7^r}{r!}\right)=0.157\) to 3 decimal places.

(b) Test the hypotheses

• \(H_0\): the number of breakdowns per day follows \(\operatorname{Po}(0.7)\);
• \(H_1\): the number of breakdowns per day does not follow \(\operatorname{Po}(0.7)\).

For a chi-squared goodness of fit test, expected frequencies should not be too small. The expected frequencies for 3, 4 and \(\geq 5\) breakdowns are combined:

Observed frequency: \(7+3+3=13\).

Expected frequency: \(5.678+0.994+0.157=6.829\), approximately.

The chi-squared test statistic is

\(\chi^2=\dfrac{(88-99.317)^2}{99.317}+\dfrac{(73-69.522)^2}{69.522}+\dfrac{(26-24.333)^2}{24.333}+\dfrac{(13-6.829)^2}{6.829}\).

So

\(\chi^2\approx 1.290+0.174+0.114+5.579=7.16\).

There are 4 combined categories and no parameter has been estimated from the data, so the degrees of freedom are \(4-1=3\).

At the 5% significance level, the critical value for \(\chi^2_3\) is \(7.815\). Since \(7.16\lt 7.815\), we do not reject \(H_0\).

Therefore, there is insufficient evidence at the 5% level to reject the expert's claim that the number of breakdowns per day follows a Poisson distribution with mean \(0.7\).