A. Two-tailed test

Example 1:

A sample of nine customers spent for lunch ($) at a fast food restaurant:

   4.20 5.03 5.86 6.45 7.38 7.54 8.46 8.47 9.87

At the 0.05 lvele of confidence, is there evidence that the mean amount spent for lunch is different from 6.50$ ?

The null hyppthesis H0 : The mean amount spent for lunch is 6.50$

The alternative hypothese H0 : The mean amount spent for lunch is different from 6.50$

spend = c(4.20, 5.03, 5.86, 6.45, 7.38, 7.54, 8.46, 8.47, 9.87)
mean(spend) # sample mean
## [1] 7.028889
sd(spend) # standard deviation
## [1] 1.812488
t.test(spend, mu = 6.50)
## 
##  One Sample t-test
## 
## data:  spend
## t = 0.87541, df = 8, p-value = 0.4069
## alternative hypothesis: true mean is not equal to 6.5
## 95 percent confidence interval:
##  5.635688 8.422090
## sample estimates:
## mean of x 
##  7.028889

The p-value = 0.4069 > 0.05. Therefore, we can not reject the null hypothesis.

The data does not provide enough evidence to conclude that the mean amount spent for lunch is different from 6.50$

Example 2:

Labels on 3.79 litre(l) cans of paint usually indicate the drying time and the area can be covered in one coat. Most brands of paint indicate that, in one coat, 3.79 l will cover between 23.2 and 46.4 square metres (m2), depending on the texture of the surface to be painted. One manufacturer, however, claims that 3.79 l of its paint wil cover 37.2 square meters of surface area. To test this claim, a random sample of ten 3.79 l cans of white paint were used to paint ten identical areas using the same kind of equipement. The actual areas (in square meters) covered by these 3.79l of paint are given here :

28.8 28.9 38.3 34.2 41.5 34.9 28.2 38.1 33.9 32.5

Do the data present sufficient evidence to indicate that the average coverage differs from 37.2 m2 ?

(Mendenhall, Introduction to Probability anf Statistic, Nelson education, p.418)

The null hypothesis H0 : mu = 37.2 The alternative hypothesis Ha : mu ≠ 37.2

sureface = c(28.8, 28.9, 38.3, 34.2, 41.5, 34.9, 28.2, 38.1, 33.9, 32.5)
mean(sureface)
## [1] 33.93
sd(sureface)
## [1] 4.488269
t.test(sureface, mu = 37.2)
## 
##  One Sample t-test
## 
## data:  sureface
## t = -2.3039, df = 9, p-value = 0.0467
## alternative hypothesis: true mean is not equal to 37.2
## 95 percent confidence interval:
##  30.71929 37.14071
## sample estimates:
## mean of x 
##     33.93

p-value = 0.034939 < 0.05.

We reject the null hypothesis.

There is sufficient evidence to indicate that the average coverage differs from 37.2 m2.

B. One-tailed test

From 58 sales, an analyst finds that the mean amount spent is 26.05$ with a standard deviation of 10.20$.

Test the hypothesis that the mean is still $24.85, as it was last year, against the alternative that it is has increased.

set.seed(0)
sale <- c(rnorm(58, mean = 26.05, sd = 10.20))
t.test(sale, mu = 24.85, alternative = c('less'), conf.level= 0.95)
## 
##  One Sample t-test
## 
## data:  sale
## t = 1.4551, df = 57, p-value = 0.9244
## alternative hypothesis: true mean is less than 24.85
## 95 percent confidence interval:
##      -Inf 28.70364
## sample estimates:
## mean of x 
##  26.64316