Statistics

1
Tutorial 9

• Statistics

2
Normal Distribution (Ex 1)

• An electrical firm manufactures light bulbs that
  have a length of life that is approximately
  normally distributed, with mean equals to 800
  hours and a standard deviation of 40 hours. Find
  the probability that random sample of 16 bulbs
  will have an average life less than 775 hours.

3
Ex 1 (cont)

• Solution
• By central limit theorem, is normally
  distributed with µ800 and
• Corresponding to 775, we find that Z
  (775-800)/10 -2.5Then,

4
Confidence Interval

• Since table for is not readily
  available,
• we transform the distribution to N(0,1) by

1-a
a/2
a/2
za/2
z1-a/2
0
5
Confidence Interval
6
Confidence Interval (Ex 2)

• An electrical firm manufactures light bulbs that
  have a length of life that is approximately
  normally distributed, with a standard deviation
  of 40 hours. If a random sample of 30 bulbs has
  an average life of 780 hours, find a 96
  confidence interval for the population mean of
  all bulbs produced by this firm.

7
Ex 2 (cont)

• Solution
• 96 gt a 0.04
• Then 96 confidence interval is hence given as
• From the table, we have Z0.98 2.055, thus

8
Significance Testing

• Step 1. State the null hypothesis, H0 and the
  alternative hypothesis.
• Step 2. Consider the appropriate distribution
  given by the null hypothesis.
• Step 3. Decide on the level of the test. This
  fixes the critical values of the test statistic.
• Step 4. Decide on the rejection criteria
• Step 5. Calculate the value of the statistic
• Step 6. Make conclusion.

9
Significance Testing (Ex 3)

• The r.v. X is such that X N(µ,100). A value is
  taken at random from the population and found to
  be 172. Test, at the 5 level, whether the
  population mean µ could be 150.

10
Ex 3 (cont)

• Solution
• Step 1 We assume that µis 150. So, H0 µ
  150 H1 µ not equal 150
• Step 2Now if H0 is true, X N(150,100)
• Step 3 4For 5 level, where Z (X -µ)/s, we
  will reject H0 if Z gt 1.96

11
Ex 3 (cont)

• Step 5.Now z (x-µ)/s (172-150)/10
  2.2
• Step 6.Since z gt 1.96, we reject H0 and
  conclude that there is significant evidence, at
  the 5 level, to suggest that the population mean
  is not 150.

12
Type I and II Error
13
Type I and II Error (Ex 4)

• To test whether a coin is fair, the following
  decision rule is adopted. Toss the coin 120
  times if the number of heads is between 50 and
  70 inclusive, accept the hypothesis that the coin
  is fair, otherwise reject it.
• a) Find the probability of rejecting the
  hypothesis when it is correct. (Type I error)
• b) With the original decision rule, find P(Type
  II error) if the coin is biased and the
  probability that a head is in fact 0.6

14
Ex 4 (cont)

• Solution
• Let X be the r.v. the number of heads obtained.
• Then XBin(n,p) with n120
• Now, since n is large, X N(np, npq)
  approximately, where q 1-p.
• Under H0, np 120x(1/2) 60
  npq 120x(1/2)x(1/2) 30So XN(60,30)
• (Remark we can estimate the r.v. by normal
  distribution when ngt30)

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Ex 4 (cont)

• a) Type I error

16
Ex 4 (cont)

• b) Type II error
• We accept H0 if the number of heads lies between
  50 and 70 inclusive.
• Now P(Type II error) P(accepting H0H1 is
  true) P(49.5 lt X lt 70.5 p 0.6)
• Now, if p0.6, np (120)(0.6) 72 npq
  (120)(0.6)(0.4) 28.8

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Ex 4 (cont)

• Under H1, XN(72, 28.8)
• Therefore, P(Type II error) 0.390 (3 sig)