Do pleasant odors help work go faster? Twenty-one subjects worked a paper-and-pe

Question

Do pleasant odors help work go faster? Twenty-one subjects worked a paper-and-pencil maze wearing a mask that was either unscented or carried the smell of flowers. Each subject worked the maze three times with each mask, in random order (hint, this is a matched pairs design). Here are the differences in their average times (in seconds), unscented minus scented. If the floral small speeds work, the difference will be positive because the time with the scent will be lower.

-7.37 -3.14 4.10 -4.40 19.47 -10.80 -.87 8.70 2.94 -17.24 14.30 -24.57 16.17 -7.84 8.60 -10.77 24.97 -4.47 11.90 -6.26 6.67

b. Find the mean and standard deviations of the 21 observations. Did the subjects work faster with the scented mask? Is the mean improvement big enough to be important?

c. Make a stemplot of the data (round to the nearest whole second). Are there outliers in the data that might hinder inference?

d. Test the hypotheses you state in (a). Is the improvement statistically significant?

Explanation / Answer

Here I write R-code for all subpart of problem as below:

b)
x=c(-7.37,-3.14,4.10,-4.40,19.47,-10.80,-.87,8.70,2.94,-17.24,14.30,-24.57,
16.17,-7.84,8.60,-10.77,24.97,-4.47,11.90,-6.26,6.67)
mean(x)
sqrt(var(x))

The output is:

> mean(x)
[1] 0.9566667 # Mean of data
> sqrt(var(x))
[1] 12.54788 #Standard deviation of data

c)
x=c(-7.37,-3.14,4.10,-4.40,19.47,-10.80,-.87,8.70,2.94,-17.24,14.30,-24.57,
16.17,-7.84,8.60,-10.77,24.97,-4.47,11.90,-6.26,6.67)
stem(x) # Stem plot of data

And output is:

> stem(x)

The decimal point is 1 digit(s) to the right of the |

-2 | 5
-1 | 711
-0 | 8764431
0 | 34799
1 | 2469
2 | 5

Thus,here "0" lie in stem part of this plot.
so we can say that there is an outlier present.

d)
x=c(-7.37,-3.14,4.10,-4.40,19.47,-10.80,-.87,8.70,2.94,-17.24,14.30,-24.57,
16.17,-7.84,8.60,-10.77,24.97,-4.47,11.90,-6.26,6.67)
t.test(x) # Testing for mean of paired data

The output is:

> t.test(x) # Testing for mean of paired data

One Sample t-test

data: x
t = 0.34938, df = 20, p-value = 0.7305
alternative hypothesis: true mean is not equal to 0
95 percent confidence interval:
-4.755061 6.668394
sample estimates:
mean of x
0.9566667

Here p-value is greater than 0.05 thus we accept null hypothesis.
And conclude that the improvement statistically not significant.

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