Question

An outbreak of Salmonella-related illness was attributed to ice cream produced at a certain factory. Scientists are interested to know whether the mean level of Salmonella in the ice cream is greater than 0.2 MPN/g. A random sample of 20 batches of ice cream was selected and the level of Salmonella measured. The levels (in MPN/g) were:

0.593, 0.142, 0.329, 0.691, 0.231, 0.793, 0.519, 0.392, 0.418, 0.219 0.684, 0.253, 0.439, 0.782, 0.333, 0.894, 0.623, 0.445, 0.521, 0.544

a)Read the data in R using a vector. Show your codes only but not the output.

b)State the two hypotheses of interest.

c) Is there evidence that the mean level of Salmonella in the icecream is greater than 0.2 MPN/g? Assume a Normal distribution and use α =0.05. Show your codes and result/output from R.

d) Interpret your finding in c)

Answer 1

Solution-A:

Salmonella_icecream <- c(0.593, 0.142, 0.329, 0.691, 0.231, 0.793, 0.519, 0.392, 0.418, 0.219, 0.684,
0.253, 0.439, 0.782, 0.333, 0.894, 0.623, 0.445, 0.521, 0.544)

Solution-B:

Null hypothesis in words:

mean level of Salmonella in the ice cream is equal to 0.2 MPN/g.

Ho:

Alternative hypothesis in words:

mean level of Salmonella in the ice cream is greater than 0.2 MPN/g.

Ha:

alpha=0.05

Solution-c:

Salmonella_icecream <- c(0.593, 0.142, 0.329, 0.691, 0.231, 0.793, 0.519, 0.392, 0.418, 0.219, 0.684,
0.253, 0.439, 0.782, 0.333, 0.894, 0.623, 0.445, 0.521, 0.544)
t.test(Salmonella_icecream,mu=0.2,alternative="greater")

Output:

   One Sample t-test

data: Salmonella_icecream
t = 6.2196, df = 19, p-value = 2.826e-06
alternative hypothesis: true mean is greater than 0.2
95 percent confidence interval:
0.4110001 Inf
sample estimates:
mean of x
0.49225
t= 6.2196

p== 2.826e-06

p<0.05

Reject null hypothesis.
Accept alternative hypothesis.

Conclusion:

There is sufficient statistical evidence at 5% level of significance to conclude that  the mean level of Salmonella in the icecream is greater than 0.2 MPN/g