Answer:
Step-by-step explanation:
given that a random sample of 50 university admissions officers was asked about expectations in application interviews.
Sample proportion = [tex]\frac{28}{50} =0.56[/tex]
[tex]H_0: p = 0.5\\H_a: p >0.5\\[/tex]
(Right tailed test at 5% significance level)
p difference = 0.06
Assuming H0 to be true p will have normal distribution with
std error = [tex]\sqrt{\frac{0.5*0.5}{50} } \\=0.0707[/tex]
Test statistic = p difference/std error
=[tex]\frac{0.06}{0.0707} \\=0.8485[/tex]
p value = 0.19808
Since p is greater than alpha, we accept null hypothesis
There is statistical evidence to support that the sample proporiton is only 0.50
Answer:
There is statistical evidence to support that the sample proportion is only 0.50.
Step-by-step explanation:
given that a random sample of 50 university admissions officers was asked about expectations in application interviews.
Sample proportion = [tex]\frac{28}{50}[/tex]
[tex]H_0:p=5[/tex]
[tex]H_0:p>5[/tex]
(Right tailed test at 5% significance level)
p difference = 0.06
Assuming H0 to be true p will have normal distribution with
std error = [tex]\sqrt{\frac{0.5\times0.5}{50} }[/tex]
Test statistic = p difference/std error
=[tex]\frac{0.06}{0.0707}[/tex]
=0.8485
p value = 0.19808
Since p is greater than alpha, we accept null hypothesis
There is statistical evidence to support that the sample proportion is only 0.50.
