Hypothesis Testing Flashcards
A drugs company claims that 75% of patients suffering from depression recover when treated with a new drug.
A random sample of 10 patients with depression is taken from a doctor’s records.
Write down a suitable distribution to model the number of patients in this sample who recover when treated with the new drug.
- X ~ B ( 10, 0.75 )
Given that the claim is correct,
Find the probability that the treatment will be successful for exactly 6 patients.
( A drugs company claims that 75% of patients suffering from depression recover when treated with a new drug.
A random sample of 10 patients with depression is taken from a doctor’s records. )
- X ~ B ( 10, 0.75 )
- P ( X = 6 ) = 0.1459980011
- = 0.1460
The doctor believes that the claim is incorrect and the percentage who will recover is lower.
From her records she took a random sample of 20 patients who had been treated with the new drug.
She found that 13 had recovered.
Stating your hypotheses clearly, test, at the 5% level of significance, the doctor’s belief.
- X ~ B ( 20, 0.75 )
- H0 : P = 0.75
- H1 : P <= 0.75
- Sig = 5% = 0.05
- P ( x <= 13 ) = 0.2142180519
- ( Binomial CD )
- 0.214 > 0.05
- So H0 is accepted, insufficient evidence to reject H0.
- Doctors belief isn’t supported by the sample
From a sample of size 20, find the greatest number of patients who need to recover from the test in part ( c ) to be significant at the 1% level.
( X ~ B ( 20, 0.75 ) )
- X ~ B ( 20, 0.75 )
- P ( X <= C ) = 0.01
- ( 1% significance level )
- Mean = 20 x 0.75 = 15
- Values <= 15
- ( Using binomial CD list function )
- 9 = 0.0039
- 10 = 0.0138
- 9 is the greatest number of patients
- ( 0.0039 < 0.01 )
A teacher thinks that 20% of the pupils in a school read the Deano comic regularly.
He chooses 20 pupils at random and finds 9 of them read the Deano.
Test, at the 5% level of significance, whether or not there is evidence that the percentage of pupils that read the Deano is different from 20%.
State your hypotheses clearly.
- X ~ B ( 20, 0.2 )
- H0 : P = 0.2
- H1 : P not = 0.2
- Sig = 5%, now 2.5% = 0.025
- P ( X => 9 ) = 1 - P ( X <= 8 ) = 1 - 0.9900182137
- ( Binomial CD )
- = 0.00998
- 0.00998 < 0.025
- H0 is rejected, the percentage of pupils that read Deano is different from 20%
*State all the possible numbers of pupils that read the Deano from a sample of size 20 that will make the test in part ( a )( i ) significant at the 5% level.
It is known from past records that 1 in 5 bowls produced in a pottery have minor defects.
To monitor production a random sample of 25 bowls was taken and the number of such bowls with defects was recorded.
Using a 5% level of significance, find critical regions for a two-tailed test of the hypothesis that 1 in 5 bowls have defects.
The probability of rejecting, in either tail, should be as close to 2.5% as possible.
- X ~ B ( 25, 1 / 5 )
- H0 : P = 0.2
- H1 : P not = 0.2
- P ( X <= x )
- Mean = 25 x 0.2 = 5
- Values <= 5
- ( Using binomial CD list function )
- 0 = 0.0037
- 1 = 0.0273
- P ( X => x ) = 1 - P ( X <= x )
- Values => 5
- 1 - 0.025 = 0.975
- ( Using binomial CD list function )
- 8 = 0.9532
- 9 = 0.9826
- ( With a 2 tailed test, choose the values closest to 2.5% )
- x <= 1
- x - 1 => 9
- x => 10
- So critical regions are, x <= 1 and x => 10
State the actual significance level of the above test.
( Lower tail, P = 0.0273 )
( Upper tail, P = 0.9826 )
- Actual sig level = 0.0273 + ( 1 - 0.9826 ) = 0.0447
- = 4.47%
At a later date, a random sample of 20 bowls was taken and 2 of them were found to have defects.
Test, at the 10% level of significance, whether or not there is evidence that the proportion of bowls with defects has decreased. State your hypotheses clearly.
( X ~ B ( 20, 0.2 ) )
- X ~ B ( 20, 0.2 )
- H0 : P = 0.2
- H1 : P < 0.2
- Sig level = 10% = 0.1
- P ( X <= 2 ) = 0.2060847189
- ( Binomial CD )
- 0.206 > 0.1
- H0 is accepted, insufficient evidence to suggest that the proportion of defective bowls has decreased.
Dhriti grows tomatoes.
Over a period of time, she has found that there is a probability 0.3 of a ripe tomato having a diameter greater than 4 cm.
She decides to try a new fertiliser.
In a random sample of 40 ripe tomatoes, 18 have a diameter greater than 4 cm.
Dhriti claims that the new fertiliser has increased the probability of a ripe tomato being greater than 4 cm in diameter.
Test Dhriti’s claim at the 5% level of significance.
State your hypotheses clearly.
- T ~ B ( 40, 0.3 )
- H0 : P = 0.3
- H1 : P > 0.3
- Sig = 5% = 0.05
- P ( X => 18 ) = 1 - P ( X <= 17 ) = 1 - 0.9680487366
- ( Binomial CD )
- = 0.0320
- 0.0320 < 0.05
- So H0 is rejected, sufficient evidence to sat that the new fertiliser has increased the probability of a ripe tomato being greater than 4 cm in diameter.
A shopkeeper knows, from past records, that 15% of customers buy an item from the display next to the till.
After a refurbishment of the shop, he takes a random sample of 30 customers and finds that only 1 customer has bought an item from the display next to the till.
Stating your hypotheses clearly, and using a 5% level of significance, test whether or not there has been a change in the proportion of customers buying an item from the display next to the till.
- X ~ B ( 30, 0.15 )
- H0 : P = 0.15
- H1 : P Not equal to 0.15
- Sig = 5%, = 0.025 ( Two tailed )
- P( X <= 1 ) = 0.0480
- ( Binomial CD )
- 0.0480 > 0.025
- H0 is accepted, insufficient evidence to reject H0, so proportion of customers buying an item from the display next to the till is 0.15.
During the refurbishment a new sandwich display was installed. Before the refurbishment 20% of customers bought sandwiches. The shopkeeper claims that the proportion of customers buying sandwiches has now increased.
He selects a random sample of 120 customers and finds that 31 of them have bought sandwiches.
Using a suitable approximation and stating your hypotheses clearly, test the shopkeeper’s claim. Use a 10% level of significance.
- X ~ B ( 120, 0.2 )
- H0 : P = 0.2
- H1 : P > 0.2
- Sig = 10% = 0.1
- u = np
- u = 120 x 0.2
- = 24
- Standard deviation = Root ( np ( 1 - p ) )
- Standard deviation = Root ( 24( 1 - 0.2 ) )
- = 4.8178049
- X ~ N ( 24, 4.8178049^2 )
- P( X => 31 ) = P( X => 30.5 )
- ( Adjusted according to normal distribution )
- P( X => 30.5 )
- ( Normal CD )
- = 0.06898205912
- 0.06898205912 < 0.1
- So H0 is rejected, proportion of customers buying sandwiches is more than 0.2
Explain what you understand by a hypothesis.
- A statement concerning a population parameter
Explain what you understand by a critical region.
- A range of values
- That would lead to the rejection of H0
In a second opinion poll of n randomly selected people it was found that no one would vote for Mrs George.
Using a 1% level of significance, find the smallest value of n for which the hypothesis H0 : p = 0.45 will be rejected in favour of H1 : p < 0.45.
( P ( X <= 5 ) = 0.05533418772 )
( Worked out from a previous question )
- Sig = 1% = 0.01
- ( 0.05533418772 )^n < 0.01
- ( Smallest value of n, so n must be less than the Sig level )
- log( 0.05533418772 )( 0.01 ) < n
- n > 7.781905532
- n > 8
- 8 is the smallest value of n
