Question

14. Students in a statistics class would like to investigate if more than two-thirds of the Earth is water.
To answer this question, students tossed an inflatable globe back and forth. After each catch, the
student recorded whether the tip of their pointer finger of their right hand was on water or land. In
50 tosses, their finger was on water 38 times. Assuming all conditions for inference are met, do the
data provide convincing statistical evidence at the significance level of a = 0.05 that more than two-
thirds of the Earth is water?
(A) Yes, because the p-value of 0.76 is greater than the significance level of 0.05.
(B) Yes, because the p-value of 0.08 is greater than the significance level of 0.05.
(C) Yes, because the p-value of 0.05 is less than the significance level of 0.08.
(D) No, because the p-value of 0.08 is greater than the significance level of 0.05.
(E) No, because the p-value of 0.03 is less than the significance level of 0.05.

Answer 1

Answer:

D

Step-by-step explanation:

No, because the p-value of 0.08 is greater than the significance level of 0.05.

Answer 2

No, there isn't convincing statistical evidence at the significance level of a = 0.05 because the p-value of 0.08 > 0.05. (Option D)

How to find the corresponding value of z test statistic for a value of proportion random variable?

Suppose the population proportion mean be denoted by $\hat{p}$

and the population proportion random variable be denoted by p

Then, the z-test statistic's value for $p = p_0$ for sample of size n is specified by

$z = \dfrac{\hat{p} - p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}$

For this case, we're specified that:

• Students in a statistics class would like to investigate if more than 2/3 of the Earth is water.
• Sample size = 50
• Favorable cases = 38
• Level of significance = 0.05

Thus, sample proportion is:

$\hat{p} = \dfrac{38}{50} = 0.76$

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses for the population proportion needs to be tested:

$\begin{array}{ccl} H_0: p & \leq & 2/3\\ H_a: p & > & 2/3 \end{array}$

This corresponds to a right-tailed test, for which a z-test for one population proportion will be used.

(2) Rejection Region

Based on the information provided, the significance level is $\alpha = 0.05$, and the critical value for a right-tailed test is $z_c = 1.64$

The rejection region for this right-tailed test is $R = \{z: z > 1.645\}$

(3) Test Statistics

The z-statistic is computed as follows:

$\begin{array}{ccl} z & = & \displaystyle \frac{\hat p - p_0}{\sqrt{ \displaystyle\frac{p_0(1-p_0)}{n}}} \\\\& = & \displaystyle \frac{0.76 - 2/3}{\sqrt{ \displaystyle\frac{ 2/3(1 -2/3)}{50}}} \\\\ & = & 1.401 \end{array}$

(4) Decision about the null hypothesis

Since it is observed that $z = 1.401 \le z_c = 1.645$, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p = 0.0806, and since $p = 0.0806 \ge 0.05$, it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis H₀ is not rejected. Therefore, there is not enough evidence to claim that the population proportion pp is greater than 2/3, at the $\alpha = 0.05$ significance level.

Thus, there isn't convincing statistical evidence at the significance level of a = 0.05 because the p-value of 0.08 > 0.05. (Option D)

Learn more about z-test for one-population proportion test here:

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