Question

A cognitive psychologist wants to know whether memory performance is changed by old age. She randomly selects 7 elderly individuals and finds that their mean score on a standardized memory test equals 418.2. Scores on the standardized test in the general population are distributed normally with a mean equal to 465.6 and a standard deviation equal to 53.6 . Is there sufficient evidence to conclude that memory performance for elderly individuals differs from that of the general population

Answer 1

Answer:

$z = \frac{418.7-465.6}{\frac{53.6}{\sqrt{7}}}= -2.340$

Now we can calculate the p value with the following probability

$p_v =2*P(z$

If we use the commonly significance level of 0.05 we see that the p value is lower than these values we can conclude that the true mean is different from the value of 465.6 at 5% of significance

Step-by-step explanation:

Data provided

$\bar X=418.7$ represent the mean score on the standardized memory test

$\sigma=53.6$ represent the population standard deviation

$n=7$ sample size    

$\mu_o =465.6$ represent the value that we want to verify

z would represent the statistic

$p_v$ represent the p value

System of hypothesis

We are interested to check if the memory performance for elderly individuals differs from that of the general population (mean different from 465.6), the system of hypothesis are then:

Null hypothesis:$\mu = 465.6$    

Alternative hypothesis:$\mu \neq 465.6$    

Since we know the population deviation we can use the z statistic from the z test for the true mean:

$z =\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$ (1)    

Replacing the info given we got:

$z = \frac{418.7-465.6}{\frac{53.6}{\sqrt{7}}}= -2.340$

Now we can calculate the p value with the following probability

$p_v =2*P(z$

If we use the commonly significance level of 0.05 we see that the p value is lower than these values we can conclude that the true mean is different from the value of 465.6 at 5% of significance