Respuesta :
Answer:
[tex]z=\frac{0.64 -0.5}{\sqrt{\frac{0.5(1-0.5)}{75}}}=2.43[/tex]
Now we can calculate the p value with the following probability:
[tex]p_v =P(z>2.43)=0.0075 \approx 0.008[/tex]
Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true proportion for this case is higher than 0.5
Step-by-step explanation:
Data given and notation
n=75 represent the random sample taken
[tex]\hat p=0.64[/tex] estimated proportion of interest
[tex]p_o=0.5[/tex] is the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level
Confidence=95% or 0.95
z would represent the statistic
[tex]p_v[/tex] represent the p value
System of hypothesis
We want to verify if the true proportion is higher than 0.5:
Null hypothesis:[tex]p =0.5[/tex]
Alternative hypothesis:[tex]p > 0.5[/tex]
The statistic is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
Replacing the info given we got:
[tex]z=\frac{0.64 -0.5}{\sqrt{\frac{0.5(1-0.5)}{75}}}=2.43[/tex]
Now we can calculate the p value with the following probability:
[tex]p_v =P(z>2.43)=0.0075 \approx 0.008[/tex]
Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true proportion for this case is higher than 0.5
Testing the hypothesis, it is found that:
- The test statistic is z = 2.42.
- The p-value is of 0.008.
- Since the p-value of the test is 0.008 < 0.05, there is significant evidence to conclude that the proportion is greater than 0.5.
The null hypothesis is:
[tex]H_0: p = 0.5[/tex]
The alternative hypothesis is:
[tex]H_0: p > 0.5[/tex].
The test statistic is given by:
[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]
In which:
- [tex]\overline{p}[/tex] is the sample proportion.
- p is the proportion tested at the null hypothesis.
- n is the sample size.
For this problem, the parameters are: [tex]\overline{p} = 0.64, p = 0.5, n = 75[/tex].
The value of the test statistic is:
[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]
[tex]z = \frac{0.64 - 0.5}{\sqrt{\frac{0.5(0.5)}{75}}}[/tex]
[tex]z = 2.42[/tex]
The p-value is the probability of finding a sample proportion above 0.64, which is 1 subtracted by the p-value of z = 2.42.
Looking at the z-table, z = 2.42 has a p-value of 0.992.
1 - 0.992 = 0.008, hence, the p-value is of 0.008.
Since the p-value of the test is 0.008 < 0.05, there is significant evidence to conclude that the proportion is greater than 0.5.
A similar problem is given at https://brainly.com/question/15350925
