Question 1
Population = Equal variance
Null hypothesis: Mean of population 1 is not higher than mean of population 2.
Alternative hypothesis: Mean of population 1 is higher than mean of population 2.
The two sample t test for equal variance
The one tailed p value comes out to be 0.0003 which is lower than significance level and hence, sufficient evidence is present to reject the null hypothesis and to accept the alternative hypothesis. Thus, it can be said that mean of population 1 is higher than mean of population 2.
Question 2
Population standard deviation = 2.25 orders per week
Sample size = 32
Null hypothesis: Average number of weekly orders is not decreased in recession period.
Alternative hypothesis: Average number of weekly orders is decreased in recession period.
It is apparent from the above that population standard deviation is given and sample size is higher than 30 and therefore, as per Central Limit Theorem the z statistic would be used to check the validity of the claim.
Average order = 17.9 orders per week
Sample mean = 15.5 orders per week
The p value = 0.00 (standard normal table)
Alpha =0.05
The p comes out to be 0.00 which is lower than significance level and hence, sufficient evidence is present to reject the null hypothesis and to accept the alternative hypothesis. Therefore, the average number of weekly orders is decreased in recession period.
Question 3
- Probability level = 0.01
Null hypothesis: Average of population 1 and population 2 is same.
Alternative hypothesis: Average of population 1 and population 2 is different.
The two sample t test for equal variance
The two tailed p value comes out to be 0.000 which is lower than significance level and hence, sufficient evidence is present to reject the null hypothesis and to accept the alternative hypothesis. Thus, it can be said that average of population 1 and population 2 is different.
- 98% confidence interval
Question 4
Margin of error would be 100
Further,
Margin of error = z value * Standard deviation / sqrt (n)
The z value for 99% confidence = 2.58
With the help of 6 sigma estimation of standard deviation
Margin of error = z value * Standard deviation / sqrt (n)
Therefore, the sample size would be 67.
Question 5
Margin of error = $2
Standard deviation = $12.50
Z value for 95% confidence = 1.96
Margin of error = z value * Standard deviation / sqrt (n)
Minimum sample size = (z value * Standard deviation/ Margin of error)^2
Margin of error = 3%
The z value for 99% confidence = 2.58
(Assuming)
As the population standard deviation is not given and thus, t stat would be used to check the claim.
Degree of freedom = 24-1 = 23
The p value (two tailed) = 0.0007
Significance level = 10% and 5%
The two tailed p value comes out to be 0.000 which is lower than significance level 5% as well as 10% and hence, sufficient evidence is present to reject the null hypothesis and to accept the alternative hypothesis. Therefore, it can be said that average weekly bill of Suburb Ever Green is more than the national average bill ($253).
Significance level = 5%
Number of observation = 24
As the population standard deviation is not given and thus, t stat would be used to check the claim.
Degree of freedom = 11-1=10
The p value (one tailed) = 0.0156
Significance level = 5%
The one tailed p value comes out to be 0.0156 which is lower than significance level 5% and hence, sufficient evidence is present to reject the null hypothesis and to accept the alternative hypothesis. Therefore, it can be seen that mean is higher than 1158.
Significance level = 2%
The one tailed p value comes out to be 0.0156 which is lower than significance level 2% and hence, sufficient evidence is present to reject the null hypothesis and to accept the alternative hypothesis. Therefore, it can be said that mean is higher than 1158.
Sample size n= 150
The p value (one tail) = 0.098
Significance level = 5% and 1%
The one tailed p value comes out to be 0.098 which is higher than significance level 5% and 1% and hence, insufficient evidence is present to reject the null hypothesis. Therefore, it can be said that proportion is not significantly lower than 79% as claimed by banking industry.
Significance level = 0.02
The one tailed p value comes out to be 0.0427 which is higher than significance level 0.02 and hence, insufficient evidence is present to reject the null hypothesis. Therefore, it can be said that per diem average expense in Adelaide has not gone up significantly.
Population standard deviation = 9.9 microgram per m^3 of air
Sample size = 37
It is apparent from the above that population standard deviation is given and sample size is higher than 30 and therefore, as per Central Limit Theorem the z statistic would be used to check the validity of the claim.
Sample mean = 75.757 microgram per m^3 of air
Standard deviation = 7.205 microgram per m^3 of air
The p value = 0.00 (standard normal table)
Alpha =0.05 and 0.01
The p comes out to be 0.00 which is lower than significance level and hence, sufficient evidence is present to reject the null hypothesis and to accept the alternative hypothesis. Therefore, it can be said that average number of lead particle of air in Adelaide City is higher than 82 microgram per m^3 of air.
Sample size n= 382
The p value (two tail) = 0.06
Significance level = 5% and 1%
The two tailed p value comes out to be 0.098 which is higher than significance level 5% and 1% and hence, insufficient evidence is present to reject the null hypothesis. Therefore, it can be said that proportion is not equal to 46.9%.
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