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Listening to music has long been thought to enhance intelligence, especially dur

ID: 3201966 • Letter: L

Question

Listening to music has long been thought to enhance intelligence, especially during infancy and childhood. To test whether this is true, a researcher records the number of hours that eight high-performing students listened to music per day for 1 week. The data are listed in the table.

(a) Find the confidence limits at a 95% CI for this one-independent sample. (Round your answers to two decimal places.)

To save money, a local charity organization wants to target its mailing requests for donations to individuals who are most supportive of its cause. They ask a sample of 5 men and 5 women to rate the importance of their cause on a scale from 1 (not important at all) to 7 (very important). The ratings for men were

M1 = 6.3.

The ratings for women were

M2 = 5.3.

If the estimated standard error for the difference

(sM1 M2)

is equal to 0.25, then consider the following.

(a) Find the confidence limits at an 80% CI for these two-independent samples. (Round your answers to two decimal places.)

4.3 4.9 5.0 3.7 4.1 5.4 4.0 4.3

Explanation / Answer

x

4.3

4.9

5

3.7

4.1

5.4

4

4.3

Mean =

4.4625

Standard deviation =

0.57802

Confidence Interval:

Here the sample size is small and the value of standard deviation is unknown we use t-distribution to find the confidence interval.

The formula for confidence interval is as below:

X bar (-/+) E

X bar = sample mean = 4.4625

E = margin of error = tc * (s / Ön)

tc is the critical value, s is the sample standard deviation and n is the sample size.

s = 0.57802

n = 8

Degrees of freedom = n – 1 = 8 – 1 = 7

We need to find critical t value from table at 5% (100% - 95%) level of significance for 7 degrees of freedom.

We get the critical value as 2.365

X bar (-/+) E

4.4652 (-/+) [ 2.365 * (0.57802 / Ö8)]

= 4.4652 (-/+) 0.4833

= 3.98 and 4.95

The 95% confidence interval is 3.98 and 4.95

x

4.3

4.9

5

3.7

4.1

5.4

4

4.3

Mean =

4.4625

Standard deviation =

0.57802

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