The average house has 12 paintings on its walls. The standard deviation is 4.7 p
ID: 3333952 • Letter: T
Question
The average house has 12 paintings on its walls. The standard deviation is 4.7 paintings. Is the mean different for houses owned by teachers? The data show the results of a survey of 14 teachers who were asked how many paintings they have in their houses. Assume that that distribution of the population is normal.
11, 15, 7, 14, 9, 12, 16, 13, 8, 14, 3, 10, 8, 9
What can be concluded at the 0.05 level of significance?
Test statistic: [ Select ] ["Z", "t", "F", "Chi-square"]
p-Value = [ Select ] ["0.042", "0.280", "0.327", "0.026"] . Round your answer to three decimal places.
[ Select ] ["Reject the null hypothesis", "Fail to reject the null hypothesis"]
Conclusion: There is [ Select ] ["insufficient", "sufficient"] evidence to make the conclusion that the population mean number of paintings that are in teacher's houses is different from 12.
UExplanation / Answer
we can conduct a 1 sample t test in R as shown below , the complete R snippet is as follow s
Alternate hypothesis is that mu is different or not equal to 12
as the number of samples is less than 40 and we want to compare the means of a sample against a value , this is a independent sample mean
data <- c(11, 15, 7, 14, 9, 12, 16, 13, 8, 14, 3, 10, 8, 9)
t.test(data,mu=12, alternative="two.sided")
The results are
> t.test(data,mu=12, alternative="two.sided")
One Sample t-test
data: data
t = -1.4075, df = 13, p-value = 0.1827 as the p value is not less than 0.05 , hence we fail to reject the null hypothesis . There is insufficient evidence to make the conclusion that the population mean number of paintings that are in teacher's houses is different from 12.
alternative hypothesis: true mean is not equal to 12
95 percent confidence interval:
8.559754 12.725960
sample estimates:
mean of x
10.64286
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