Outline

Question 1

1.  -  = the difference between the mean number of the duration (in days) of the cold symptoms if 25 volunteers take zinc lozenge every two or three waking hours until their cold symptoms disappeared and 23 volunteers take placebo lozenge every two or three waking hours until their cold symptoms disappeared.[pic 1][pic 2]

1. :  -  = 0 (no zinc lozenge effect)[pic 3][pic 4][pic 5]

1. :  -  ≠ 0 (a zinc lozenge affect exist; 2-sided hypothesis)[pic 6][pic 7][pic 8]

1.  -  = 8.1 – 4.5 = 3.6[pic 9][pic 10]

1. −test statistic[pic 11]

Use  = [pic 12][pic 13]

Estimate = 3.6       hypothesised value = 0

Se (-) =  =  = 0.493[pic 14][pic 15][pic 16][pic 17]

df = minimum() = minimum (24,22) = 22[pic 18]

 =  = 7.302[pic 19][pic 20]

1. P-value  pr() + pr() (2-tiales test) = 2.59596E-07 (from excel)[pic 21][pic 22][pic 23]

1. P- value interpretation :

We have very strong evidence:

▪ against  in favour of  that a zinc lozenge affect exists for the mean number of the duration (in days) of the cold symptoms[pic 24][pic 25]

▪ the mean number of the duration (in days) of the cold symptoms for the zinc lozenge volunteers is not the same as for placebo lozenge volunteers.

1. 95% CI for  -  :[pic 26][pic 27]

Use estimate  t  se(estimate)[pic 28][pic 29]

Estimate = 3.6            se (-) = 0.493[pic 30][pic 31]

t-multiplier = 2.074

confidence interval = 3.6 ± 2.074×0.493 = (2.58, 4.62)

1. CI interpretation:

With 95% confidence, we estimate the mean number of the duration (in days) of the cold symptoms for the volunteers take placebo lozenge every two or three waking hours until their cold symptoms disappeared is somewhere between 2.6 and 4.6 days longer than the mean number of the duration of the cold symptoms for volunteers take zinc lozenge every two or three waking hours until their cold symptoms disappeared.

Question 2

1. Situation B (single sample – New Zealanders aged 18 to 34; different response categories – agree or disagree with having a new flag for New Zealand)

1.  = the proportion of New Zealanders aged 18 to 54 who agree that New Zealand should adopt a new flag =  = 0.1839[pic 32][pic 33]

 = the proportion of New Zealanders aged 55 years and over who agree that New Zealand should adopt a new flag =  = 0.2786[pic 34][pic 35]

 -  = the difference between the proportion of New Zealanders aged 55 years and over who agree that New Zealand should adopt a new flag and the proportion of New Zealanders aged 18 to 54 who agree that New Zealand should adopt a new flag.[pic 36][pic 37]

1. :  -  = 0[pic 38][pic 39][pic 40]

1. :  -  ≠ 0[pic 41][pic 42][pic 43]

1.  -  = 0.2786 – 0.1839 = 0.0947[pic 44][pic 45]

1. T- test statistic:

Use  =  [pic 46][pic 47]

Estimate = 0.0947              hypothesised value = 0

Se ( - ) : situation A[pic 48][pic 49]

Se ( - ) =  =  = 0.03767             df= [pic 50][pic 51][pic 52][pic 53][pic 54]

 =  = 2.513937[pic 55][pic 56]

1. P-value   pr() + pr() (2-tiales test) =0.011939 (from excel)[pic 57][pic 58][pic 59]

1. P-value interpretation:

We have strong evidence:

▪ against  in favour of  that there is a difference between the proportion of New Zealanders aged 55 years and over who agree that New Zealand should adopt a new flag and the proportion of New Zealanders aged 18 to 54 who agree that New Zealand should adopt a new flag.[pic 60][pic 61]

1. 95% CI for  - :[pic 62][pic 63]

Use estimate  t  se(estimate)[pic 64][pic 65]

Estimate = 0.0947           Se ( - )  = 0.03767[pic 66][pic 67]

t-multiplier = 1.960

confidence interval = 0.947 ± 1.960×0.03767 = (0.873, 1.021)

1. Cl interpretation:

With 95% confidence, we estimate the proportion of New Zealanders aged 55 years and over who agree that New Zealand should adopt a new flag is somewhere between 0.87 and 1.02 higher than the proportion of New Zealanders aged 18 to 54 who agree that New Zealand should adopt a new flag.

1. Not sure, because the true proportions cannot be calculated.

Qusetion3

1. Hypothesised value   = 0         ≠ 0[pic 68][pic 69]

 =  =  = -1.05 left direction[pic 70][pic 71][pic 72]

1. Case 2 and case 4

1. Case 2 has practical significance. Case 1,4 and 5 don’t have practical significance.

1. Case 1 and case 3

1. With 95% confidence, we estimate the compressive strength of concrete for paving using method 1 is somewhere between 502 psi and 578 psi higher than compressive strength of concrete for paving using method 2. The 540 psi difference between the compressive strength of concrete for paving using method 1 and 2 is statistically significant (P-value = 0.0000) but we do not have sufficient information to be able to determine whether the true difference between method 1 and 2 has practical significance. I will recommend method 1 because with 95% confidence we can say that the difference will be greater than 500 psi which will result in considerable benefit for the factory.

Qusetion4

1. There is a cluster, no outliers, slight skewness and the sample size is 12.
2. T-TEST PAIRS=wetsuit WITH nowetsuit (PAIRED)

  /CRITERIA=CI(.9500)

  /MISSING=ANALYSIS.

T-Test