Chapter 10, Problem 70CR

Students at the Akademia Podlaka conducted an experiment to determine whether the Belgium-minted Euro coin was equally likely to land heads up or tails up. Coins were spun on a smooth surface, and in 250 spins, 140 landed with the heads side up (New Scientist, January 4, 2002).

1. a. Should the students interpret this result as convincing evidence that the proportion of the time the coin would land heads up is not 0.5? Test the relevant hypotheses using α = 0.01.
2. b. Would your conclusion be different if a significance level of 0.05 had been used? Explain.

To determine

Check whether the sample data provide convincing evidence that the proportion of time a coin landed as heads up is different from 0.5 at 0.01 significance level.

Answer to Problem 70CR

No, the sample data do not provide the evidence that the proportion of time the coin landed as heads up is different from 0.5.

Explanation of Solution

Calculation:

In a sample of 250 spinning of Belgium-minted Euro coin, 140 spins were founded to turn up as heads.

Step 1:

Population characteristic of interest:

Let p be the population characteristic of interest.

In this context, p denotes the population proportion of spins that landed as heads up.

Step 2:

Null hypothesis: $H_0:p=0.5$

That is, the proportion of spins that landed as heads up is 0.5.

Step 3:

Alternative hypothesis: $H_a:p\ne 0.5$

That is, the proportion of spins that landed as heads up is different from 0.5.

Step 4:

Significance level, $\alpha$:

It is given that the significance level, $\alpha =0.01$.

Step 5:

Test statistic, z:

$z=\frac{\widehat{p}-p}{\sqrt{\frac{p\left(1-p\right)}{n}}}$,

Where, $\widehat{p}$ be the sample proportion, p is the hypothesized proportion, and n is the sample size.

Substitute the hypothesized proportion, $p=0.5$, in the test statistic.

$z=\frac{\widehat{p}-0.5}{\sqrt{\frac{0.5\left(1-0.5\right)}{n}}}$

Here, the sample proportion $\widehat{p}$ is not known.

Step 6:

Assumptions:

• Let $\widehat{p}$ be the sample proportion from a random sample.
• The large sample z test can be used if the sample size n satisfies the conditions: $n\left(\text{hypothesized value}\right)\ge 10$ and $n\left(1-\text{hypothesized value}\right)\ge 10$.
• The sample size should not be greater than 10% of the population size.

Requirement check:

• It is assumed that the sample is from a random sample.
• Check the conditions: $n\left(\text{hypothesized value}\right)\ge 10$ and $n\left(1-\text{hypothesized value}\right)\ge 10$.

$\begin{array}{c}n\left(\text{hypothesized value}\right)=np\\ =250\left(0.5\right)\\ =175\\ >10\end{array}$

$\begin{array}{c}n\left(1-\text{hypothesized value}\right)=n\left(1-p\right)\\ =250\left(1-0.5\right)\\ =250\left(0.5\right)\\ =175\\ >10\end{array}$

Since $n\left(\text{hypothesized value}\right)$ and $n\left(1-\text{hypothesized value}\right)$ are greater than 10, the sample size requirement is met.

Therefore, the large sample z test is appropriate.

• Although, the population size is not known, it is reasonable to assume that the sample size of 250 acts as the representative for all the spins of Euro coins. It is also definite that the sample size is less than the 10% of the population of the spins of Euro coins.

Step 7:

The value of the test statistic is obtained as follows:

$z=\frac{\widehat{p}-0.5}{\sqrt{\frac{0.5\left(1-0.5\right)}{n}}}$

The sample proportion, $\widehat{p}$, is obtained as follows:

$\widehat{p}=\frac{x}{n}$

Where, x is the number of observations of interest and n is the sample size.

Substitute the corresponding values to get the sample proportion.

$\begin{array}{c}\widehat{p}=\frac{140}{250}\\ =0.56\end{array}$

By substituting the value of sample proportion, the test statistic is obtained as follows:

$\begin{array}{c}z=\frac{0.56-0.5}{\sqrt{\frac{0.5\left(1-0.5\right)}{250}}}\\ =\frac{0.06}{0.032}\\ \approx 1.90\end{array}$

Thus, the value of test statistic is 1.90.

Step 8:

P-value:

In this context, the alternative hypothesis denotes that the test carried out is two-tailed test. Therefore, the P-value is the area under the z curve and the two-tailed of the calculated z value.

The P-value for the test statistic value of 1.90 is obtained as follows:

$\begin{array}{c}P\text{-value}=2\times \left\{\text{1}-\text{Area to the Right of }1.90\right\}\\ =2\times \left\{1-P\left(z<1.90\right)\right\}\end{array}$

Use standard normal probabilities (Cumulative z curve areas) table to find the z-value.

Procedure:

For z at 1.90:

         Locate 1.9 in the left column of the table.

         Obtain the value in the corresponding row below .00.

That is, $P\left(z<1.90\right)=0.9713$.

The probability of the event is as given below:

$\begin{array}{c}P\text{-value}=2\left\{1-P\left(z<1.90\right)\right\}\\ =2\left\{1-0.9713\right\}\\ =2\left\{0.0287\right\}\\ =0.0574\end{array}$

Thus, the P-value for the test statistic of 1.90 is 0.0574.

Step 9:

Decision rule:

If $P\text{-value}\le \text{Significance level}$, then reject the null hypothesis $H_0$.

If $P\text{-value}>\text{Significance level}$, then fail to reject the null hypothesis $H_0$.

Here, the $P\text{-value}$ of 0.0574 is greater than the significance level 0.01.

That is, $P\text{-value}\left(=0.0574\right)>\text{Significance level}\left(=0.01\right)$.

The decision is that the null hypothesis is not rejected.

Conclusion:

Hence, the sample data do not provide the evidence that the proportion of time the coin landed as heads up is different from 0.5.

To determine

Explain whether the conclusion for the hypothesis test is different at significance level $\alpha =0.05$.

Explanation of Solution

Increasing significance level increases, the chance to reject the null hypothesis.

It is given that the significance level $\alpha =0.05$.

Here, the $P\text{-value}$ of 0.0574 is less than the significance level 0.05.

That is, $P\text{-value}\left(=0.0574\right)\text{> Significance level}\left(=0.05\right)$.

Therefore, it can be concluded that the null hypothesis is not rejected at the 0.05 significance level.

Hence, there is no evidence that the proportion of time the coin landed as heads up is different from 0.5.

In this context, even if the significance is increased, there is no change in the conclusion.