Maths Study Material: Confidence Interval, Hypothesis Testing, Non-linear Regression, Forecasting Model

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PART A
Question 1
99% confidence interval
Samples size = 8
Mean of sample = 244.5
Standard deviation of sample = 21.7
Here, population standard deviation is not given and also, the sample size is lower than 30
and hence, as per the central limit theorem the t test would be taken into account.
Degree of freedom = Sample size – 1 = 8 -1 = 7
The t stat for 99% confidence interval and 7 degree of freedom = 3.4995
Standard error = Standard deviation
√¿ ¿
Margin of error = t value * standard error = 3.4995 * 7.6721 = 26.848
Now,
Lower limit of 99% confidence interval = Mean - Margin of error = 244.5 – 26.848 = 217.65
Upper limit of 99% confidence interval = Mean + Margin of error =244.5 +26.848 = 271.35
99% confidence interval = [ 217.65 271.35]
It can be said with 99% confidence that the sample mean i.e. number of calories in 3 ounces
of French fries would fall within the range of 217.65 and 271.35.
Question 2
Claim: Female adults blood cell count is lower than normal i.e. 4.8.
Step 1: Null and alternative hypotheses
Null hypothesis H0 : μ≥ 4.8
Alternative hypothesis Ha : μ<4.8
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Step 2: Test statistic
Samples size = 6
Mean of sample = 4.4
Standard deviation of sample = 0.28
Here, population standard deviation is not given and also, the sample size is lower than 30
and hence, as per the central limit theorem the t test would be taken into account.
t= Mean−hypothesized mean
Standard deviation
√ sample ¿ ¿= 4.4−4.8
0.28
√6
=−3.4993 ¿
¿
Step 3: The p value
It is a left tailed hypothesis.
Degree of freedom = Sample size – 1 = 6 -1 = 5
The p value corresponding to degree of freedom and t stat = 0.0086
Step 4: Significance level
Significance level (alpha) = 0.05
Step 5: Conclusion
It is apparent that the p value is lower than significance level and therefore, sufficient
evidence is present to reject the null hypothesis and to accept the alternative hypothesis.
Therefore, it can be concluded that the red blood cell count of the patient is lower than
normal i.e. 4.8.
PART B
Question 1
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Claim: Whether there is a statistically significant difference in reading age of the two groups.
Step 1: Null and alternative hypotheses
Null hypothesis H0 : μD=0
Alternative hypothesis Ha : μD ≠ 0
Step 2: Test statistic
Samples size = 6
Difference in sample mean = -2.33 months
Difference in sample standard deviation = 2.16 months
Here, population standard deviation is not given and also, the sample size is lower than 30
and hence, as per the central limit theorem the t test would be taken into account.
t= Difference∈sample mean
Difference∈sample standard deviation
√sample ¿ ¿=−2.33
2.16
√ 6
=−2.64 ¿
¿
Step 3: The p value
It is a two tailed hypothesis test.
Degree of freedom = Sample size – 1 = 6 -1 = 5
The p value corresponding to degree of freedom and t stat = 0.046
Step 4: Significance level
Significance level (alpha) = 0.05
Step 5: Conclusion
It is apparent that the p value (0.046) is lower than significance level (0.05) and therefore,
sufficient evidence is present to reject the null hypothesis and to accept the alternative
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