It is recommended that adults get 8 hours of sleep each night. A researcher hypothesized college students got less than the recommended number of hours of sleep each​ night, on average. The researcher randomly sampled 20 college students and obtained a​ p-value of 0.10. Suppose the researcher sampled more college students and that the sample mean and sample standard deviation stayed the same. Would the​ p-value be​ lower, be​ higher, or stay the​ same?

Respuesta :

Answer:

Sampling more College students by the research will lower the p-value.

Step-by-step explanation:

A larger sample size provides consistent evidence of the non zero effect against the null hypothesis than a smaller sample size. A sample size of 20 students collected by the researcher is prone to random error and bias compare to a larger sample size. A p-value closer to 0 indicates strong evidence against a null hypothesis.

Answer:

The p-value becomes lower.

Step-by-step explanation:

Let [tex]\mu[/tex] (unknown) be the true mean of number of hours of sleep each night. If we have the hypothesis [tex]H_{0}: \mu = \mu_{0}[/tex] vs [tex]H_{1}: \mu < \mu_{0}[/tex], the test statistic for a small sample size n = 20 is [tex]T=\frac{\bar{X}-\mu_{0}}{S/\sqrt{n}}[/tex], where  [tex]\bar{X}[/tex] and S are the sample mean and standard deviation respectively. If [tex]H_{0}[/tex] is true, we know that T has a t distribution with n-1 degrees of freedom. The p-value for the lower-tail alternative is computed as [tex]P(T < t_{1})[/tex], where [tex]t_{1}[/tex] is the observed value for the given sample. If we increase the size of the sample and the sample mean and sample standard deviation stayed the same, then, the value [tex]s/\sqrt{n}[/tex] becomes smaller, besides, this quantity is always positive. The absolute value of [tex]\bar{x}-\mu_{0}[/tex] stays the same, and the absolute value of the observed value [tex]t_{2}=\frac{\bar{x}-\mu_{0}}{s/\sqrt{n}}[/tex] becomes bigger.  

If [tex]t_{1}[/tex] is negative, then, [tex]t_{2}[/tex] is also negative and smaller than [tex]t_{1}[/tex], therefore, [tex]P(T < t_{2})[/tex] becomes lower.