Question

. One such theory hypothesizes that people should spontaneously follow a

24-hour cycle of sleeping and waking-even if they are not exposed to the usual pattern of

sunlight. To test this notion, 8 paid volunteers were placed in a room in which there was no light from the outside and no clocks or other indicators of time. After a month in the room, each individual developed a cycle. Their cycles at the end of the study were as follows: 25, 27, 25, 23, 24, 25, 26, 25. Using the .05 significance level, determine if these participants’ cycles are different from a 24-hour cycle.

SS = 10,

Sample mean = 25

1.what is the null hypothesis?

2.what is the research hypothesis?

3.What is the comparison mean?

4.what is the comparison distribution?

5.What is/are the cutoff(s)?

6.what is the t score for the sample score?

7.How is the denominator of a t-score different from the denominator of a z-score?

8.what is your conclusion for this hypothesis test?

Answer 1

1.

The null hypothesis to be tested in this question is,

$H_0:\mu=24$

2.

The research hypothesis is the alternative hypothesis and it is given as,

$H_1:\mu\not=24$

3.

The comparison mean is the sample mean $\bar{x}$ and its value is given as,

$\bar{x}=\sum(x)/n$, where $n=8$ and $\sum(x)=25+27+25+23+24+25+26+25=200$

Therefore,

$\bar{x}=200/8=25$

4.

The comparison distribution used is the student's t distribution. It is used because the population variance$(\mu)$ is unknown and the sample size $(n=8)$ is considerably small.

5.

The cutoffs are the t-table values that leaves an area of $\alpha/2=0.05/2=0.025$ each to the right and to the left with $n-1=8-1=7$ degrees of freedom. The values are given as,

$t_{0.025,7}=2.365$ and $-t_{0.025,7}=-2.365$ since the t distribution is symmetric about the mean.

Therefore, the cut off points are, $(-2.365,\space 2.365)$

6.

The t-score for the sample is given as,

$t^*=(\bar{x}-\mu)/(s/\sqrt{n})$

In order to determine the sample standard deviation $(s)$ , we first evaluate for the variance given as,

$V(x)=\sum(x-\bar{x})^2/(n-1)=10/7=1.428571429$

Now,

$s=\sqrt{V(x)}=\sqrt{1.428571429}=1.1952(4\space dp)$

Thus $t^*=(25-24)/(1.1952/\sqrt{8})=1/0.422577127=2.3664(4\space dp)$

The t score for the sample score is $t^*=2.3664$

7.

The denominator of both scores is the standard error and it is given as,

$\sigma/\sqrt{n}$ for a Z-score and $s/\sqrt{n}$ for a t-score. The main difference is that the standard error $(s/\sqrt{n})$ for a t-score is smaller than the standard error$(\sigma/\sqrt{n})$ for a Z-score.

8.

For this hypothesis test, the null hypothesis is rejected if $|t^*|\gt t_{0.025,7}$

Since $|t^*|=|2.3664|=2.3664\gt t_{0.025,7}=2.365$, we reject the null hypothesis and conclude that sufficient evidence exist to show that  participants’ cycles are different from a 24-hour cycle at 5% level of significance.