QUESTION
Answer1. (a) The distribution Y would be also a normal distribution, because it is composed of 10 normal distributions. The mean of this new distribution would be 4620 g and the variance would be 10 times the variance of the box distribution that is 250 g.
(b)
Now the normal distribution with Z value equal to -1.4, would have the area to the left of the mean=0.0808.So, the probability that the mean weight of cereal box contents in a randomly selected carton is less than 455g is (0.5-0.0808)=0.4192 .
Answer2. (a) The box plot of the given data has been plotted in the excel sheet on the tab Question 2. As can be easily seen from the plot the distribution is symmetric.
2(b)The Values of the mean, s and the critical value of t are shown in the excel tab TtestQ2.The Formula for
Where the first mean is the sample mean, and the second mean is the population means, s is the standard deviation and n is the degrees of freedom.
2. (c) The boxplot shows that the data is equally distributed about the median, which can be also confirmed from the data given, by the location of the median as calculated in the sheet.
2. (d) Yes, In order to get the maximum profits can be done if the policies are within the 95% confidence interval.
Answer 3.(a)Except the maximum value which is $990440, the other data are located in the range of $ 100 difference only
(b) As derived in the sample statistic:
| p – One sided |
0.361759 |
Accept Null Hypothesis because p > 0.05 (Means are the same) |
| p – two sided |
0.723518 |
Accept Null Hypothesis because p > 0.05 (Means are the same) |
(c) The formula for test statistic is
The equation is given in symbolic form in the answer of part 2(b).
(d)The sampling distribution is shown in the excel sheet attached. We have assumed a 95% level of confidence for this distribution to reject the null hypothesis.
(e) If the level of significance is 5%, the level of confidence is 95%. The critical value of the t statistic is 0.362
(f)
| Test Mean |
478000 |
| Confidence Level |
0.95 |
| N |
13 |
| Average | $461,335 |
| Stdev |
165908 |
| SE Mean |
46014.59 |
|
|
|
|
|
(g) The observed value of the test statistic is T=0.362
| p – One sided |
0.361759 |
| p – two sided |
0.723518 |
(h) The p value for the test at 5% significance level is 0361. the null hypothesis.
Gives: Accept Null Hypothesis because p > 0.05
When it is two tailed the null hypothesis gives:
Accept Null Hypothesis because p > 0.05
The tells us about the house prices the probability to get a cheaper house is 0.361
LA36
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