Sampling distribution

Sampling Distribution
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Chapter 9
Question 9.7
This is probability distribution of means that is obtained from a specific selected random samples of size, n from a particular population
Question 9.8
• Population mean is equal to sample mean
• The shape of the curve is normal
• The standard deviation divided by the sample size n is equal to the standard errors of the population.
Question 9.9
a) True
b) True
c) False
d) True
e) True
f) False
Question 9.13
a) Standard error= SD/(√Sample Size
SD=1*√36
=standard deviation is 6
b) 2= sd/√36
Standard deviation= 12
c) 5=SD/√36 =6*5
=5*6
Standard deviation =30
d) 100=SD/√36 = 6*100
Standard deviation=600
Question 9.14
a) In this case, we can make an assumption and conclude that the 144 children is our random sample and is adequate to give a normal curve.
b) This is an indication that majority of the reported hours for the sampling distribution will be below 8 hours or be above the mean of the random sample.
c) Based on the sample mean of 21 and standard deviation of 8 the sample data will be below 13 hours and higher to 29 hours.
d) 10 hours
e) 5 and 37 hours
Chapter 10
Question 10.9
Null hypothesis:
Ha: u=21.75
There is no significant difference between the BMI and the body size.
For alternative hypothesis, it states that there is a significant difference between BMI and the body size.
If p< significance level (=0.01) we fail to accept the null hypothesis at a significance level of 0.

01 the p-value is 0.433. From the values, it is evident that the p-value is greater than the significant value. Therefore, we will accept the null hypothesis and then conclude that the results are statistically significant.
Question 10.10
When the level of significance is 0.05, the p-value is 0.2. It is evident that the p-value is less than the level of significance, and, therefore, we reject the null hypothesis and conclude that there is significant difference between BMI and the body size.
Question 10.11
Test
Mean is not equal to 35
Null hypothesis: u=35
This is a two-tailed test
Mean=30
Standard deviation=10
Sample size= 20
Significance level= 5%
Degree of freedom n-1=19
=x-u/s√n
=30-35/10/√20=-2.236
The test statistic is -2.2336
Using the t-table to read the p-values which are 0.02 and 0.05
T-critical t0.25, 19=2.093
T<-2.093 and t>2.093
In this case, we reject the null hypothesis because the p-value is less than the critical value and conclude that at 0.05 significance level we have adequate information to say that u is different from 35.
Question 10.12
When the level of significance is 5%, the p-value is 0. 013.The p-value is higher than the significance level. In this situation, we fail to accept the null hypothesis and conclude that average commuting time in Chicago area is different as compared to the US census reported.
Chapter 11
Question 11.11
Any hypothesis cannot be tested directly due to lack of precision of the testing instruments. Therefore, it is essential to specify a certain number in which the sampling distribution will be based. Additionally, the hypothesis cannot be tested directly because we have the alternative hypothesis in which we construct our arguments and a test should be done to make a strong case (Robert, 2004).
Question 11.19
a) Type I error occurs when we reject the null hypothesis when it is true, and this happens when a high significance level is used. Therefore, to reduce the chances of this error, we should use a lower significance level.
b) Type II error occurs when we fail to reject the null hypothesis when it is still false. It can be controlled by changing sample size used in the test (Robert, 2004).
Question 11.20

0.2
0.9
0.5
Chapter 12
Question 12.7
In this case, we shift to a point with a level of confidence which is lesser such as 90 and 75.
The sample size should be increased.
Question 12.8
We have
)  ,    and n=36
The 95% confidence interval for mean is

; Z-value from sd. normal table

The mean of the random sample lies between 32.99 and 33.09 with a 5% level of significance. In my opinion yes, the producer can manufacture candy boxes with an average exceeding 32.
Question 12.10
3 (102, 106)
1 (100,102). This is because the error margin is small in this section.
5 (91,98)
4 (90,111). In this case, the confidence interval is high as compared to the sample variance.

Reference
Robert, C. P. (2004). Monte carlo methods. John Wiley & Sons, Ltd.