Respuesta :
Answer: No, it does not seem to be a valid estimate if in a random sample of 90 college students, 28 are found to ride bicycles to class.
Step-by-step explanation:
Since we have given that
Number of college students = 90 = n
Number of students ride bicycles to class = 28 = x
So, [tex]\hat{p}=\dfrac{x}{n}=\dfrac{28}{90}=0.31[/tex]
Hypothesis :
[tex]H_0:p=\hat{p}\\\\H_a:\hat{p}\leq p[/tex]
So, the test statistic value would be
[tex]\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}\\\\z=\dfrac{0.31-0.25}{\sqrt{\dfrac{0.25\times 0.75}{90}}}\\\\z=\dfrac{0.06}{0.0456}\\\\z=1.136[/tex]
At 0.05 level of significance,
critical value z = 1.96
Since 1.96>1.136
So, we will accept the null hypothesis.
Hence, No, it does not seem to be a valid estimate if in a random sample of 90 college students, 28 are found to ride bicycles to class.
No, it does not seem to be a valid estimate if in a random sample of 90 college students, 28 are found to ride bicycles to class.
What is level significance?
The level of significance is the measurement of the statistical significance. It defines whether the null hypothesis is assumed to be accepted or rejected. It is expected to identify if the result is statistically significant for the null hypothesis to be false or rejected.
Given that:
Number of college students, n = 90
Number of students ride bicycles to class, x = 28
So, p°= x/n
= 28/90
=0.31
Now, Hypothesis:
H[tex]_0[/tex]: p°=p
H[tex]_a[/tex]: p°≤p
Thus, the statistical value will be
=[p°-p/√p(1-p)]/n
= 0.31-0.25/√0.25*0.75]/90
=1.136
At 0.05 level of significance,
z = 1.96
and, 1.96>1.136
So, we will accept the null hypothesis.
Hence, it does not seem to be a valid estimate.
Learn more about level of significance here:
https://brainly.com/question/9163177
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