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 4.3Explanation / 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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