SOLUTION: The average house has 15 paintings on its walls. Is the mean different for houses owned by teachers? The data show the results of a survey of 15 teachers who were asked how many p
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Question 1197672: The average house has 15 paintings on its walls. Is the mean different for houses owned by teachers? The data show the results of a survey of 15 teachers who were asked how many paintings they have in their houses. Assume that the distribution of the population is normal.
15, 13, 15, 15, 14, 12, 12, 14, 14, 15, 14, 15, 16, 13, 13
What can be concluded at the
α
= 0.05 level of significance?
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!
Answer: The number of paintings that teachers have is different from 15 paintings.
===============================================================
Explanation:
= Greek letter mu, represents the population mean
More specifically, mu is the population mean of the number of paintings per teacher household.
Null Hypothesis: mu = 15
Alternative Hypothesis:
The "not equal" sign in the alternative says we have a two-tailed test.
The data set is
{15, 13, 15, 15, 14, 12, 12, 14, 14, 15, 14, 15, 16, 13, 13}
There are n = 15 items in this data set.
Add up all of the values in the data set to get
15+13+15+15+14+12+12+14+14+15+14+15+16+13+13 = 210
Divide that sum by n = 15 to compute the mean
mean = (sum of values)/n
mean = (210)/15
mean = 14
This is the sample mean xbar. Its job is to estimate the population mean mu.
Use a graphing calculator, spreadsheet, or statistical software to compute the standard deviation of the data set. We want the sample standard deviation.
You should find the sample standard deviation is approximately s = 1.195229
Let's compute the standard error (SE)
SE = s/sqrt(n)
SE = 1.195229/sqrt(15)
SE = 0.3086068007936
SE = 0.308607
This value is approximate.
Now we can find the test statistic
t = (xbar - mu)/SE
t = (14 - 15)/0.308607
t = -3.24036719841092
t = -3.24
The sample size is n = 15
The degrees of freedom (df) is
df = n-1
df = 15-1
df = 14
Use a stats calculator, spreadsheet, or similar to find the p-value is roughly 0.0059
Since we're doing a two-tailed test, we find the area to the left of t = -3.24 and then double that result
The p-value being smaller than alpha = 0.05 tells us to reject the null and conclude that is the case.
Therefore we conclude that the number of paintings owned by teachers is different from 15 paintings.
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