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University personnel are concerned about the sleeping habits of students and the negative impact on academic performance. In a random sample of 377 U.S. college students, 209 students reported experiencing excessive daytime sleepiness (EDS). Is there sufficient evidence to conclude that more than half of U.S. college students experience EDS? Use a 5% level of significance.

Respuesta :

Answer: The claim is false.

Step-by-step explanation:

Since we have given that

n = 377

x = 209

We will use one sample proportion test:

As we know that

[tex]\hat{p}=\dfrac{x}{n}=\dfrac{209}{377}=0.55[/tex]

Since we need to check whether more than half of U.S. college students experience EDS.

So, hypothesis are as follows:

[tex]H_0:p\leq 0.5\\\\H_1:p>0.5[/tex]

So, it becomes

[tex]test\ statistic=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}\\test\ statistic=\dfrac{0.55-0.5}{\sqrt{\dfrac{0.5(1-0.5)}{377}}}\\test\ statistic=1.94[/tex]

According to z-table, we get that

P-value = 0.9738

As we can see that

0.97>0.05

So, we accept the null hypothesis.

Hence, the claim is false.

Answer:

The claim of reporting of 209 students experiencing excessive day time sleepiness is false.

Step-by-step explanation:

Given information:

Sample size [tex]n=377[/tex]

Student reported [tex]x=209[/tex]

Now if we use the random proportion test:

[tex]p=x/n\\p=209/377\\p=0.55[/tex]

Now, check if more then half of the college students experience EDS.

So, the hypothesis are:

[tex]H_0: p\leq 0.5\\H_1:p>0.5.\\[/tex]

Now. we need to find the [tex]p[/tex] value according to Z-table :

As we have:

[tex]p=0.9738[/tex] from the Z-table

0.975>0.55

S, one can accept the null hypothesis.

Hence the claim is false.

So, The claim of reporting of 209 students experiencing excessive day time sleepiness is false.

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