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Hypothesis Testing Flashcards

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1
Q

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.

A
  • X ~ B ( 10, 0.75 )
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2
Q

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. )

A
  • X ~ B ( 10, 0.75 )
  • P ( X = 6 ) = 0.1459980011
  • = 0.1460
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3
Q

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.

A
  • 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
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4
Q

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 ) )

A
  • 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 )
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5
Q

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.

A
  • 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%
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6
Q

*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.

A
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7
Q

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.

A
  • 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
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8
Q

State the actual significance level of the above test.
( Lower tail, P = 0.0273 )
( Upper tail, P = 0.9826 )

A
  • Actual sig level = 0.0273 + ( 1 - 0.9826 ) = 0.0447

- = 4.47%

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9
Q

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 ) )

A
  • 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.
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10
Q

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.

A
  • 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.
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11
Q

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.

A
  • 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.
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12
Q

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.

A
  • 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
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13
Q

Explain what you understand by a hypothesis.

A
  • A statement concerning a population parameter
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14
Q

Explain what you understand by a critical region.

A
  • A range of values

- That would lead to the rejection of H0

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15
Q

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 )

A
  • 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
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16
Q

State the conditions under which the normal distribution may be used as an approximation to the binomial distribution.

Study These Flashcards
A
  • P is close to 0.5

- N is large

17
Q

A random sample of size n is to be taken from a population that is normally distributed with mean 40 and standard deviation 3.
Find the minimum sample size such that the probability of the sample mean being greater than 42 is less than 5%.

Study These Flashcards
A
  • X ~ N ( 40, 3^2 / n )
  • P ( X > 42 ) < 0.05
  • ( Inverse Normal )
  • Area = 0.05
  • Standard deviation = 1
  • u = 0
  • Z = 1.644853667
  • ( Positive due to the < arrow )
  • Z = X - u / ( Standard deviation / Root ( n ) )
  • 1.644853667 = 42 - 40 / ( 3 / Root ( n ) )
  • 1.644853667 ( 3 / Root ( n ) ) = 42 - 40
  • 4.934561001 / Root ( n ) = 2
  • Root ( n ) / 4.934561001 = 1 / 2
  • Root ( n ) = 2.467280501
  • n => 6.087473068
  • ( => because P ( X > 42 ) )
  • n = 7
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