91影视

The provided histogram shows the digits drawn in California鈥檚 win fourlotteries.

a.

A deceptive graph is a visual representation of data thatcreates a disparity from the true set of observations.

The graph does not start from 0, but from 15.

The resultant histogram creates a deception that there is a large disparity between the counts of the categories.

Therefore, it can be concluded that the scaling on the vertical axis is wrong in the graph.

b.

The shape of the histogram shows the pattern followed by the observations in the data. The shape can be indicative of the appropriate distribution that underlies the observations.

For example:

• If the shape is almost bell-like,the underlying distribution is normal for the observations.
• If the shape is constant, forming a rectangle parallel to the x-axis,the underlying distribution is uniformfor the observations.

As per the description, the shape of the histogram is not bell-like.Thus, the distribution is non-normal distribution.

As the distribution of digits from 0 to 9 is expected to be constant (same probability for each digit), the shape of the histogram is expected to be a flat curve (rectangular shape of the same height).

c.

The frequencies are equivalent to the heights of the columns corresponding to each digit.

Observing the histogram, you get the frequencies as listed below:

The hypothesis to test the claim that that the digits are selected from a population in which the digits are all equally likely are given below:

\(\begin{array}{l}{H_o}:{p_0} = {p_1} = {p_2} = ... = {p_9}\\{H_a}:{\rm{at}}\;{\rm{least}}\;{\rm{one}}\;{\rm{proportion}}\;{\rm{is}}\;{\rm{different}}\end{array}\)

Here,\({p_i}\)is the proportion of frequencies for the i-th digit.

Assume the significance level is 0.05.

The test statistic is

\({\chi ^2} = \sum {\frac{{{{\left( {O - E} \right)}^2}}}{E}} \).

Here, O is the observed frequency and E is the expected frequency.

The sum of frequencies is 200.

For uniform distribution, the expected frequencies are the same in number at\(\frac{{200}}{{10}} = 20\).

Thus, the test statistic is\({\chi ^2} = 3.000\).

The degree of freedom is calculated below:

\(\begin{array}{c}df = n - 1\\ = 10 - 1\\ = 9\end{array}\)

The p-value is computed using the chi-square table as shown below:

\(\begin{array}{c}p - value = P\left( {{\chi ^2} > 3.000} \right)\\ = 0.965\end{array}\)

As the p-value exceeds the significance measure of 0.05, the null hypothesis fails to be rejected.

Thus, there is enough evidence to prove that the digits have the same frequencies and the distribution of digits is uniform.

There appears to be no problem with the lottery as there is enough evidence to prove that there are even chances of getting any digit since they are uniformly distributed.