The average person aged 15 or older gets 8 hours and 23 minutes (503 minutes) of

Question

The average person aged 15 or older gets 8 hours and 23 minutes (503 minutes) of sleep per night. To test if this average has changed recently, a random sample of 53 people aged 15 years or older was selected, and the number of minutes they slept recorded Assume the standard deviation of hours of sleep is 64 minutes. Using = 0.20, complete parts a through c below a. Explain how Type l and Type ll errors can occur in this hypothesis test. A Type l error can occur when the researcher concludes the average hours of sleep | but the the average hours of sleep A Type ll error can occur when the researcher concludes that the average hours of sleep fact, the average hours of sleep | when, in

Explanation / Answer

a.
i.
Given that,
Standard deviation, =64
Sample Mean, X =500
Null, H0: =504
Alternate, H1: !=504
Level of significance, = 0.2
From Standard normal table, Z /2 =1.282
Since our test is two-tailed
Reject Ho, if Zo < -1.282 OR if Zo > 1.282
Reject Ho if (x-504)/64/(n) < -1.282 OR if (x-504)/64/(n) > 1.282
Reject Ho if x < 504-82.048/(n) OR if x > 504-82.048/(n)
-----------------------------------------------------------------------------------------------------
Suppose the size of the sample is n = 53 then the critical region
becomes,
Reject Ho if x < 504-82.048/(53) OR if x > 504+82.048/(53)
Reject Ho if x < 492.73 OR if x > 515.27
Suppose the true mean is 500
Probability of Type I error,
P(Type I error) = P(Reject Ho | Ho is true )
= P(492.73 < x OR x >515.27 | 1 = 500)
= P(492.73-500/64/(53) < x - / /n OR x - / /n >515.27-500/64/(53)
= P(-0.827 < Z OR Z >1.737 )
= P( Z <-0.827) + P( Z > 1.737)
= 0.2041 + 0.0412 [ Using Z Table ]
= 0.245

ii.
Given that,
Standard deviation, =64
Sample Mean, X =500
Null, H0: =504
Alternate, H1: !=504
Level of significance, = 0.2
From Standard normal table, Z /2 =1.2816
Since our test is two-tailed
Reject Ho, if Zo < -1.2816 OR if Zo > 1.2816
Reject Ho if (x-504)/64/(n) < -1.2816 OR if (x-504)/64/(n) > 1.2816
Reject Ho if x < 504-82.022/(n) OR if x > 504-82.022/(n)
-----------------------------------------------------------------------------------------------------
Suppose the size of the sample is n = 53 then the critical region
becomes,
Reject Ho if x < 504-82.022/(53) OR if x > 504+82.022/(53)
Reject Ho if x < 492.733 OR if x > 515.267
Implies, don't reject Ho if 492.733 x 515.267
Suppose the true mean is 500
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(492.733 x 515.267 | 1 = 500)
= P(492.733-500/64/(53) x - / /n 515.267-500/64/(53)
= P(-0.827 Z 1.737 )
= P( Z 1.737) - P( Z -0.827)
= 0.9588 - 0.2041 [ Using Z Table ]
= 0.755
For n =53 the probability of Type II error is 0.755

b.
Given that,
Standard deviation, =64
Sample Mean, X =500
Null, H0: =504
Alternate, H1: !=504
Level of significance, = 0.2
From Standard normal table, Z /2 =1.2816
Since our test is two-tailed
Reject Ho, if Zo < -1.2816 OR if Zo > 1.2816
Reject Ho if (x-504)/64/(n) < -1.2816 OR if (x-504)/64/(n) > 1.2816
Reject Ho if x < 504-82.022/(n) OR if x > 504-82.022/(n)
-----------------------------------------------------------------------------------------------------
Suppose the size of the sample is n = 53 then the critical region
becomes,
Reject Ho if x < 504-82.022/(53) OR if x > 504+82.022/(53)
Reject Ho if x < 492.733 OR if x > 515.267
Implies, don't reject Ho if 492.733 x 515.267
Suppose the true mean is 506
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(492.733 x 515.267 | 1 = 506)
= P(492.733-506/64/(53) x - / /n 515.267-506/64/(53)
= P(-1.509 Z 1.054 )
= P( Z 1.054) - P( Z -1.509)
= 0.8541 - 0.0656 [ Using Z Table ]
= 0.789
For n =53 the probability of Type II error is 0.789

c.
Given that,
Standard deviation, =64
Sample Mean, X =500
Null, H0: =504
Alternate, H1: !=504
Level of significance, = 0.2
From Standard normal table, Z /2 =1.2816
Since our test is two-tailed
Reject Ho, if Zo < -1.2816 OR if Zo > 1.2816
Reject Ho if (x-504)/64/(n) < -1.2816 OR if (x-504)/64/(n) > 1.2816
Reject Ho if x < 504-82.022/(n) OR if x > 504-82.022/(n)
-----------------------------------------------------------------------------------------------------
Suppose the size of the sample is n = 53 then the critical region
becomes,
Reject Ho if x < 504-82.022/(53) OR if x > 504+82.022/(53)
Reject Ho if x < 492.733 OR if x > 515.267
Implies, don't reject Ho if 492.733 x 515.267
Suppose the true mean is 498
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(492.733 x 515.267 | 1 = 498)
= P(492.733-498/64/(53) x - / /n 515.267-498/64/(53)
= P(-0.599 Z 1.964 )
= P( Z 1.964) - P( Z -0.599)
= 0.9752 - 0.2746 [ Using Z Table ]
= 0.701
For n =53 the probability of Type II error is 0.701

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