Respuesta :
Using the z-distribution, it is found that the correct option is:
b. is not significantly greater than 50 percent.
At the null hypothesis, it is tested if the proportion is not significantly greater than 50%, that is:
[tex]H_{0} : P\leq 0.5[/tex]
At the alternative hypothesis, it is tested if the proportion is significantly greater than 50%, that is:
[tex]H_{1} : P > 0.5[/tex]
The test statistic is given by
[tex]Z=\frac{P-p}{\sqrt{\frac{p(1-p)}{n} } }[/tex]
In which:
[tex]P[/tex] is the sample proportion.
[tex]p[/tex] is the proportion tested at the null hypothesis.
n is the sample size.
For this problem, the parameters are: .
Hence:
Z=[tex]\frac{0.55-0.5}{\sqrt{\frac{0.5(1-0.5)}{100} } }[/tex]
Z=1
The critical value for a right-tailed test, as we are testing if the mean is greater than a value, using the z-distribution with a significance level of 0.05, is of z^* = 1.645
Since the test statistic is less than the critical value for the right-tailed test, there is not enough evidence to conclude that the proportion is greater than 50%, hence, option b is correct.
learn more about of z-distribution here
https://brainly.com/question/16207101
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