Hypothesis testing

1
Hypothesis testing

• Established records show that the lengths of
  mussels from an estuary are normally distributed
  with a mean of 30 mm. A random sample of 25
  mussels was taken from a polluted beach.
• Is there sufficient evidence to conclude that
  pollution appears to inhibit the growth of
  mussels?

2
Data

• 27.72 17.44 19.72 42.39 22.31
• 30.87 20.06 18.03 16.29 24.95
• 19.15 32.22 27.33 35.88 18.57
• 22.02 27.45 26.56 22.32 31.40
• 19.12 43.56 40.63 36.12 26.95

3
Display the data
4
Summarise the data

• N Mean Median StDev
• 25 26.76 26.56 8.10
• Minimum Maximum Q1 Q3
• 16.29 43.56 19.43 31.81

5
One-sample t-test

• Null hypothesis pollution has no effect.
• Mean length has not changed.
• H0 ? 30
• Alternative hypothesis pollution inhibits growth
• Mean length is less than previously
• H1 ? lt 30

6
One or two tailed test

• One tailed tests
• alternative hypothesis H1 ? lt 30
• alternative hypothesis H1 ? gt 30
• Two tailed tests
• alternative hypothesis H1 ? ? 30

7
Test statistic

• Assuming the data are taken from a normal
  distribution then t follows a Student's t
  distribution with n-1 degrees of freedom.

8
Calculations
9
P-value

• What is the probability of obtaining a mean of
  26.76 or less if the true mean is 30?
• This is known as the p-value.
• If p is small then the null hypothesis is
  rejected.
• 5 significance level. If p lt0.05 the reject H0.

10
24 degrees of freedomTable 5
11
Table 5
P 15 10 5
2.5 1 0.5 0.1 0.05   n
1 1.963 3.078 6.314 12.706 31.820
63.655 318.275 636.438 2
1.386 1.886 2.920 4.303 6.965 9.925
22.327 31.596 3 1.250 1.638
2.353 3.182 4.541 5.841 10.214
12.923 4 1.190 1.533 2.132
2.776 3.747 4.604 7.173 8.610
5 1.156 1.476 2.015 2.571 3.365
4.032 5.893 6.869   21
1.063 1.323 1.721 2.080 2.518 2.831
3.527 3.819 22 1.061 1.321
1.717 2.074 2.508 2.819 3.505
3.792 23 1.060 1.319 1.714
2.069 2.500 2.807 3.485 3.768
24 1.059 1.318 1.711 2.064 2.492
2.797 3.467 3.745 25
1.058 1.316 1.708 2.060 2.485 2.787
3.450 3.725
12
StatgtBasic statisticsgt1sample t
13
Output

• Test of mu 30.00 vs mu lt 30.00
• Variable N Mean StDev SE Mean
  T P
• length 25 26.76 8.10 1.62
  -2.00 0.029
• Note that tables are not needed to interpret this
  output
• Since p 0.029 lt 0.05 then H0 can be rejected

14
Conclusion

• Always state the final conclusion in terms of the
  original problem and avoiding statistical jargon.
• There is evidence to suggest that pollution does
  inhibit the growth of the mussels.

15
Test of normality
16
95 confidence interval

• What is the average length of mussels in the
  polluted area?
• Point estimate - 26.76
• Confidence interval

17
Critical value for 2.5
18
Table 5
P 15 10 5
2.5 1 0.5 0.1 0.05   n
1 1.963 3.078 6.314 12.706 31.820
63.655 318.275 636.438 2
1.386 1.886 2.920 4.303 6.965 9.925
22.327 31.596 3 1.250 1.638
2.353 3.182 4.541 5.841 10.214
12.923 4 1.190 1.533 2.132
2.776 3.747 4.604 7.173 8.610
5 1.156 1.476 2.015 2.571 3.365
4.032 5.893 6.869   21
1.063 1.323 1.721 2.080 2.518 2.831
3.527 3.819 22 1.061 1.321
1.717 2.074 2.508 2.819 3.505
3.792 23 1.060 1.319 1.714
2.069 2.500 2.807 3.485 3.768
24 1.059 1.318 1.711 2.064 2.492
2.797 3.467 3.745 25
1.058 1.316 1.708 2.060 2.485 2.787
3.450 3.725
19
Calculation
20
Note

• A 95 confidence interval is equivalent to using
  a 5 significance level for a two-tailed test.
• A higher level of confidence will result in a
  wider interval.

21
1 in 20
m
22
1 in 20
m
23
1 in 20
m
24
1 in 20
m
25
1 in 20
m
26
1 in 20
m