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kap26 [50]
4 years ago
14

Dr. Park is interested in how the functioning of the brain changes with age and how this, in turn, affects older individuals’ me

mory abilities. She randomly selects 11 older and 16 younger adults to study pictures of landscapes presented for 5 seconds each. After a 10-minute rest period, both groups perform a visual recognition test. They view a series of pictures and categorize each as either "old" (part of the set they studied) or "new" pictures. While this test is occurring, researchers use fMRI to measure the participants’ brain activity. Dr. Park and her team notice that the older adults all produced similar amounts of activity during the task. The younger adults, on the other hand, showed much more diversity of brain activity. Some younger study participants had a large amount of activity, while others had much less. Would it be valid for Dr. Park to use the independent-measures t test to test whether the brain activity of younger adults is different from that of older adults during a visual recognition task?
a. Yes, because none of the assumptions of the independent-measures t test are violated.
b. Yes, because the independent-measures t test is a robust test. No, because the two groups studied are not independent.
c. No, because the two populations from which the samples are selected do not appear to have equal variances.
Mathematics
2 answers:
a_sh-v [17]4 years ago
8 0

Answer:

Option C.  No, because the two populations from which the samples are selected do not appear to have equal variances.

Step-by-step explanation:

The variance measures how far a set of (random) numbers are spread out from their average value.

The fact that the younger adults show diversity in their brain activity while the older adults produce similar activities show that there are no equal variances in the two populations from which the samples are selected.

Therefore it would not be valid for Dr. Park to use the independent-measures t-test to test whether the brain activity of younger adults is different from that of older adults during a visual recognition task

AnnZ [28]4 years ago
5 0

Answer:

No, because the two populations from which the samples are selected do not appear to have equal variances.

Step-by-step explanation:

We know that in the given study the two groups studied are older and younger people. The samples are randomly selected and are independent of each other. The t-test is a robust test if the assumptions are followed. In the given study, we are given that the older adults produced a similar response, while the younger adults produced more of diverse brain activity. Therefore, the variance of younger adults would not be equal to the variance of older adults. Since one of the assumptions of the t-test is that the population of the two groups have same variance and here this assumption appears to be violated. Hence the correct option is:  No, because the two populations from which the samples are selected do not appear to have equal variances.

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Step-by-step explanation:

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Why there must be at least two lines on any given plane
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3 years ago
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A random sample of size n1= 25, taken from a normal population with a standard deviation σ1= 5, has a mean X1= 80. A second rand
sesenic [268]

Answer:

The 94% confidence interval would be given by 2.898 \leq \mu_1 -\mu_2 \leq 7.102

Step-by-step explanation:

Previous concepts  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

\bar X_1 =80 represent the sample mean 1

\bar X_2 =75 represent the sample mean 2

n1=25 represent the sample 1 size  

n2=36 represent the sample 2 size  

\sigma_1 =5 sample standard deviation for sample 1

\sigma_2 =3 sample standard deviation for sample 2

\mu_1 -\mu_2 parameter of interest.

The confidence interval for the difference of means is given by the following formula:  

(\bar X_1 -\bar X_2) \pm z_{\alpha/2}\sqrt{\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2}} (1)  

The point of estimate for \mu_1 -\mu_2 is just given by:

\bar X_1 -\bar X_2 =80-75=5

Since the Confidence is 0.94 or 94%, the value of \alpha=0.06 and \alpha/2 =0.03, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.03,0,1)".And we see that z_{\alpha/2}=1.88  

The standard error is given by the following formula:

SE=\sqrt{\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2}}

And replacing we have:

SE=\sqrt{\frac{5^2}{25}+\frac{3^2}{36}}=1.118

Confidence interval

Now we have everything in order to replace into formula (1):  

5-1.88\sqrt{\frac{5^2}{25}+\frac{3^2}{36}}=2.898  

5+1.8\sqrt{\frac{6^2}{36}+\frac{7^}{49}}=7.102  

So on this case the 94% confidence interval would be given by 2.898 \leq \mu_1 -\mu_2 \leq 7.102  

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Answer:

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2x = 180-38

2x = 142

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