Question

it is common knowledge that a fair coin lands heads up 50% of the time and tails up 50% of the time. for fair coins, a probability of 0 is assigned to the coin landing on its edge. but what if several coins were glued together? a teacher claims that if 10 pennies were glued together and flipped many times, the penny stack would land on heads 25% of the time, on tails 25% of the time, and on its edge 50% of the time. to investigate this claim, a student glues 10 pennies together and flips the stack 200 times. the student finds that the stack lands on heads 42 times, tails 41 times, and its edge 117 times. the student would like to know if there is convincing evidence that the distribution differs from what the teacher claimed. what is the value of the chi - square test statistic and p - value?
find the chi - square table here.
○ ( chi^{2}=5.79 ), p - value is between 0.05 and 0.10
○ ( chi^{2}=5.79 ), p - value is between 0.10 and 0.15
○ ( chi^{2}=5.97 ), p - value is between 0.025 and 0.05
○ ( chi^{2}=5.97 ), p - value is between 0.10 and 0.15

Explanation:

Step1: Calculate expected frequencies

The teacher claims \(P(\text{heads}) = 0.25\), \(P(\text{tails})=0.25\), \(P(\text{edge}) = 0.5\). The number of trials \(n = 200\).
The expected frequency formula is \(E_i=n\times p_i\).
For heads: \(E_{\text{heads}}=200\times0.25 = 50\)
For tails: \(E_{\text{tails}}=200\times0.25 = 50\)
For edge: \(E_{\text{edge}}=200\times0.5=100\)

Step2: Calculate the chi - square test statistic

The chi - square test statistic formula is \(\chi^{2}=\sum\frac{(O_i - E_i)^{2}}{E_i}\), where \(O_i\) is the observed frequency.
For heads: \(\frac{(42 - 50)^{2}}{50}=\frac{(- 8)^{2}}{50}=\frac{64}{50}=1.28\)
For tails: \(\frac{(41 - 50)^{2}}{50}=\frac{(-9)^{2}}{50}=\frac{81}{50}=1.62\)
For edge: \(\frac{(117 - 100)^{2}}{100}=\frac{17^{2}}{100}=\frac{289}{100}=2.89\)
\(\chi^{2}=1.28 + 1.62+2.89=5.79\)

Step3: Determine the degrees of freedom and find the P - value

The degrees of freedom formula for a goodness - of - fit test is \(df=k - 1\), where \(k\) is the number of categories. Here \(k = 3\), so \(df=3 - 1=2\)
Looking at the chi - square table with \(df = 2\):
\(\chi^{2}_{0.10,2}=4.605\) and \(\chi^{2}_{0.05,2}=5.991\)
Since \(4.605<5.79<5.991\), the \(P\) - value is between \(0.05\) and \(0.10\)

Answer:

\(\chi^{2}=5.79\), \(P\) - value is between \(0.05\) and \(0.10\) (the first option)