Answer:
a. The Z-statistic is [tex]Z = \frac{\hat{p}-0.2}{\sqrt{0.2(0.8)/200}}[/tex] and the observed value 1.591
b. p-value: 0.0558
Step-by-step explanation:
To see if there is evidence that the psychic is doing better than just guessing, we want to test the next hypothesis
[tex]H_{0}: p = 0.2[/tex] vs [tex]H_{a}: p > 0.2[/tex] (upper-tail alternative)
We have a large sample size of n = 200 trials. Therefore, the test statistic is given by
[tex]Z = \frac{\hat{p}-0.2}{\sqrt{0.2(0.8)/200}}[/tex] which is normally distributed.
The observed value is Â
[tex]z = \frac{49/200-0.2}{\sqrt{0.2(0.8)/200}} = 1.591[/tex]. Â
The p-value is calculated as P(Z > 1.591) = 0.0558. With this p-value we fail to reject the null hypothesis at the significance level of 0.05 for instance.
Based on the calculations, the z-statistic for this test is equal to 1.59 and the p-value for this test is 0.0559.
A null hypothesis ([tex]H_0[/tex]) can be defined the opposite of an alternate hypothesis ([tex]H_a[/tex]) and it asserts that two (2) possibilities are the same.
For the null hypothesis, we would test that:
[tex]H_o : p =0.2[/tex]
For the alternate hypothesis, we would test that:
[tex]H_o > p =0.2[/tex]
For the sample portion, we have:
[tex]\bar{p}=\frac{x}{n} \\\\\bar{p}=\frac{49}{200}\\\\\bar{p}=0.245[/tex]
For the standard deviation of the sample portion, we have:
[tex]\delta_p = \sqrt{\frac{0.2(1-0.2)}{200} } \\\\\delta_p =0.0283[/tex]
The test statistics would be calculated with this formula:
[tex]Z_o=\frac{\bar{p}\;-\;0.2}{ \delta _p }\\\\Z_o=\frac{0.245\;-\;0.2}{0.0283 }[/tex]
Zo = 1.59.
For the p-value:
Since it is an upper tailed test, the p-value is given by:
[tex]P(Z > Z_o)=P(Z > 1.59)=P(Z < -1.59)=0.0559[/tex]
From the z-table, a z-score of 1.59 corresponds has a p-value of 0.0559. Therefore, the p-value for this test is 0.0559 > 0.05.
In conclusion, we fail to reject the null hypothesis because [tex]0.0559 \geq 0.05[/tex]
Read more on null hypothesis here: https://brainly.com/question/14913351
