Suppose a test of musical ability has a normal distribution with a mean of 50 and a standard deviation of 5. Approximately what percentage of people are over 55?
a. 6%
b. 16%
c. 45%
d. 84%
A z-score of 1.75 indicates how many standard deviation units above the mean?
a. 1.75
b. 3.5
c. 4.75
d. 2.25
If you have a confidence interval of 112 to 136,
a. All of the subjects scores fall within this range
b. You have an effect size of 12
c. You can be 95% sure that this interval includes the true population mean
d. You can be 95% sure that the results did not happen by chance.
If the significance level for a two-tailed test is .05, what is the corresponding significance level for each tail of the test?
a. .05
b. .025
c. .075
d. .1
Z score is your score in terms of standard deviations.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions desired.
If you are talking about a 95% confidence interval, C.
.05/2 = ?
To find the percentage of people who are over 55 on the musical ability test, we need to calculate the area under the normal distribution curve above the value of 55. Here's how you can do it:
1. Start by calculating the z-score for a value of 55 using the formula:
z = (x - μ) / σ
where x is the value (55 in this case), μ is the mean (50), and σ is the standard deviation (5).
z = (55 - 50) / 5
z = 1
2. Once you have the z-score, you can look up the corresponding area under the standard normal distribution curve using a z-table or a calculator. Since we want the area above 55, we are interested in the cumulative probability from 1 to positive infinity (tail area).
A z-score of 1 corresponds to an area of approximately 0.8413.
3. To find the percentage, multiply the area by 100:
Percentage = 0.8413 * 100
Percentage ≈ 84.13%
Therefore, the percentage of people who are over 55 is approximately 84%. The correct answer is option d. 84%.
Now, let's move on to the next question.
The z-score measures how many standard deviation units above or below the mean a particular value is. In this question, we need to find how many standard deviation units above the mean a z-score of 1.75 represents.
To calculate this, we can use the formula:
z = (x - μ) / σ
Given that we know the z-score (1.75), but we need to find x, which represents the number of standard deviation units above the mean.
1. Rearrange the formula to solve for x:
x = z * σ + μ
2. Plug in the values:
x = 1.75 * 5 + 50
x = 8.75 + 50
x = 58.75
Therefore, a z-score of 1.75 indicates that the value is 1.75 standard deviation units above the mean. The correct answer is none of the given options.
Moving on to the third question.
A confidence interval provides a range of values within which the true population mean is likely to fall. In this case, we have a confidence interval of 112 to 136.
1. The lower bound of the confidence interval is 112, which means we can be confident that the true population mean is at least 112.
2. The upper bound of the confidence interval is 136, which means we can be confident that the true population mean is at most 136.
3. The width of the confidence interval is determined by subtracting the lower bound from the upper bound: 136 - 112 = 24.
Therefore, with a confidence interval of 112 to 136, none of the given options accurately describe it.
Now, let's move on to the last question.
The significance level, denoted by α ("alpha"), determines the probability of rejecting the null hypothesis when it is true. For a two-tailed test with a significance level of 0.05, we need to divide this overall significance level between the two tails of the distribution equally.
Since each tail corresponds to half of the overall significance level, the corresponding significance level for each tail of the test is 0.05 divided by 2.
Therefore, the correct answer is option b. 0.025.