Basic Practice Of Statistics Tasks 2022

BASIC PRACTICE OF
STATISTICS
STUDENT ID:
[Pick the date]

Task 1
Checking Assumptions
Since there is no explicit mention of the underlying sampling, hence an SRS has been
assumed. Further, the histogram of the given data has been obtained in order to check if the
distribution of the sample data can be assumed to be normal or not.
Based on the above histogram, it can be assumed that the underlying data is normally
distributed since it is almost symmetric. Additionally, central limit theorem also states that if
the underlying sample size is greater than 30, then the underlying sample can be assumed as
normally distributed. Since, the sample size is greater than 30, normality of sample can also
be established from this theorem.
Further, it is noteworthy that the population standard deviation has been given which implies
that the appropriate test statistics would be z and not t. T is a suitable test statistic when
population standard deviation is unknown and also the sample size is small. Hence, one
sample z test would be used here.
TESTING HYPOTHESIS
The sample mean has been determined using Excel functions for average. Sample mean =
11.8875
Standard error = 3/5000.5 = 0.1342
The requisite hypotheses for the given problem are mentioned below.
Null hypothesis H0 :μ=11 i.e. average stem length of rose is 11.
Alternative hypothesis Ha : μ ≠11 i. e . stem length of roseis not 11.

Z= x −μ0
s / √n
Putting the input values obtained above, we get z = (11.8875-11)/0.1342 = 6.15
The corresponding p value for the above z value is 0.
The requisite 95% confidence interval for the mean has been computed using Excel and
relevant screenshot is provided below.
Hence, the requisite confidence interval is (11.639,12.136).
CONCLUSION
Since p value is less than assumed significance level of 1%, hence the null hypothesis would
be rejected and alternative hypothesis would be accepted. Hence, the conclusion is that mean
stem lengths of roses is different from 11. Since the confidence interval does not include 11,
hence the average length of stem of roses for population exceeds 11.
Task 2
Checking Assumptions
The given distribution is binomial considering that there are only two alternatives to choose
from i.e. Yes and NO. Also, each of the responses are independent from each other. In order
to test hypothesis, we need to test whether the underlying binomial distribution can be
considered as normal distribution.
For the above to happen, following conditions ought to be satisfied.
np ≥ 5
np(1-p) ≥ 5

where n = number of trials
and p = probability of success
In the given case, n = 500, p = 0.84
n*p = 500*0.84 = 420
np(1-p) = 500*84*0.16 = 67.2
Clearly, both the above terms are greater than 5 owing to which the given binomial
distribution can be approximated as normal distribution. Thus, the appropriate test statistics
for testing of hypothesis would be Z.
TESTING HYPOTHESIS
The requisite hypotheses are as stated below.
Null Hypothesis: p = 0.8 i.e. only 80% of the population like dogs
Alternative Hypothesis: p > 0.8 i.e. more than 80% of the population like dogs
The hypothesis testing can be performed using the test statistic and also using the confidence
interval approach. Both have been carried out below.
The requisite formula to be used is given below.
The above computations have been performed in Excel and respective p value has been
found. It has come out as 0.0127. The relevant screenshot is shown below.