STA1010 Statistical Methods For Science

STA1010 Statistical Methods For Science

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STA1010 Statistical Methods For Science

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STA1010 Statistical Methods For Science

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Course Code: STA1010
University: Monash University is not sponsored or endorsed by this college or university

Country: Australia

a) What percent of eggs have weights above 70 grams?b) What percent have scores between 70 and 75 grams?c) What is the lowest weight of the heaviest 5% of eggs?d) Eggs are sold in packs of one dozen. What is the distribution of the weight of packs?e) What proportion of packs has total weight above 800 grams? 
Probability that drug cause side effect to patient p =3/100 = 0.03
Total patient n = 5
Here, the distribution would be binomial distribution.  

Probability that none of the five patients experience side effects

Probability that at least two patients experience side effects
(c) Total patient n = 5
Here, the distribution would be Poisson distribution.
? =np = 5(0.03) = 0.15 

Probability that none of the five patients experience side effects

Probability that at least two patients experience side effects

It can be said that probabilities computed with the help of binomial distribution is lesser than Poisson distribution. Further, the probabilities found in part (b) are  more accurate than computed in part (c). 

Standard deviation
Normal distribution

% of eggs have weight above 70 grams P(X>70) 

Hence, 15.92%  of eggs have weight above 70 grams. 

% of eggs have scores between 70  and 75 grams 

Hence, 13.59% of eggs have scores between 70  and 75 grams. 

Lowest weight of heaviest  5% egg

The p value = 1- 0.05 = 0.95
Z score for P (0.95) =1.645  

Distribution of weight of packs

Each pack has an SRS size of 12
Normal distribution   

Proportion of packs has total weight above 800 grams

Standard deviation = sqrt(300) = 17.32 
Therefore, 0.1251 proportions of packs has total weight above 800 grams. 
For 9 to 12, the average value has been taken = (9+10+11+12)/4 = 10.5
Mean = 3.81
Variance = 18.1875- mean ^2 = 18.18 – (3.81)^2 = 3.6714 
(b) Sample size = 100
It can be seen that sample size is higher than 30 and hence, it can be said based on central limit theorem that the x bar can be approximately by a normal distribution. 
Mean = (1/100)*
Standard deviation = sqrt (3.6714) = 1.914
S.E. = 1.91/ Sqrt(100) = 0.1914
(c) Sample size = 100 
There is a 0.8394 probability that the speed measurement would be between 3 and 4 on the Beaufort scale. 

The data is termed as paired data. 

Summary statistics 

Mean change 
Standard deviation  

95% confidence interval

Level of significance = 0.05
The t value
Lower limit of 95% confidence interval
Upper limit of 95% confidence interval
Therefore, the 95% confidence interval [1.39    10.01] mmHg

Hypotheses testing 

It can be seen from the above that 0 value does not fall in the 95% confidence interval and hence, sufficient evidences are present to reject the null hypothesis and to accept the alternative hypothesis at 5% confidence interval. Thus, it can be concluded that systolic blood pressure has fallen an hour after the visit of the dentist. 

Sample size needs to be determined.

Sample size = (Z value* Standard deviation /margin of error)^2
The z value for 95% confidence = 1.96
Standard deviation = 5 mmHg
Margin of error = 1
Sample size = (1.96* 5 /1)^2 = 96.04
Therefore, the sample size should be equal to or greater than 97 in order to achieve the mean change in systolic blood pressure to an accuracy of +/-1 mmHg.  

There are two situation sleep or no sleep which means it is not a paired situation. 

& (c) Summary statistic

For sleep 
Mean = 218/10 = 21.8  
Standard deviation = sqrt(206.509/10) = 4.6
For no sleep 
Mean = 218/9 = 19.22
Standard deviation = sqrt(280.12/9) = 5.74
Pooled standard deviation = sqrt(((8*4.614^2) + (7*5.74^2))/15) =5.17
Null and alternative hypotheses
H0: No difference between the scores of the two groups
H1: difference between the scores of the two groups
The test stat (t value) = ((21.8-19.22)/5.17)* sqrt(1/9 + 1/8) = 1.03
Degree of freedom = 15 
The p value = 0.316
It can be seen that p value is higher than level of significance and hence, null hypotheses would not be rejected. Therefore, no difference is present between the scores of the two groups.

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