Exercise 5.1: Inferential Statistics
Q1
Identify the below statement as biased or Unbiased statement. Justify your answer.
"For a survey about daily mobile uses by students, random selection of twenty students from a school"
"For a survey about daily mobile uses by students, random selection of twenty students from a school"
Answer: Unbiased
Reasoning: The statement specifies a "random selection" of students. In statistics, Simple Random Sampling is considered an unbiased method because every member of the population (students in the school) has an equal probability of being selected. This eliminates selection bias.
Q2
(i) Find the critical \( t \) value for \( \alpha = 0.01 \) with d.f.= 22 for a left-tailed test.
(ii) Find the critical \( t \) values for \( \alpha = 0.10 \) with d.f.=18 for a two-tailed \( t \) test.
(ii) Find the critical \( t \) values for \( \alpha = 0.10 \) with d.f.=18 for a two-tailed \( t \) test.
Answer: (i) -2.508 (ii) ±1.734
- Degree of freedom (d.f.) = 22
- Significance level \( \alpha = 0.01 \)
- Looking at the t-table for \( \text{d.f.} = 22 \) and one-tail area \( 0.01 \), the value is \( 2.508 \).
- Since it is left-tailed, the critical value is negative: -2.508.
- Degree of freedom (d.f.) = 18
- Significance level \( \alpha = 0.10 \)
- For a two-tailed test, we look up \( \alpha/2 = 0.05 \) in each tail.
- Looking at the t-table for \( \text{d.f.} = 18 \) and tail area \( 0.05 \), the value is \( 1.734 \).
- Critical values are +1.734 and -1.734.
Q3
Suppose that a 95% confidence interval states that population mean is greater than 100 and less than 300. How would you interpret this statement?
Answer:
We are 95% confident that the true population mean lies between the interval of 100 and 300.
Note: This implies that if we were to take many random samples and calculate a confidence interval for each, approximately 95% of those intervals would contain the actual population mean.
Q4
A shoe maker company produces a specific model of shoes having 15 months average lifetime. One of the employees in their R & D division claims to have developed a product that lasts longer. This latest product was worn by 30 people and lasted on average for 17 months. The variability of the original shoe is estimated based on the standard deviation of the new group which is 5.5 months. Is the designer's claim of a better shoe supported by the findings of the trial? Make your decision using two tailed testing using a level of significance of \( p < .05 \).
Answer: No. Null Hypothesis Accepted.
1. Hypotheses:
Null Hypothesis (\( H_0 \)): \( \mu = 15 \) (No change in lifetime)
Alternative Hypothesis (\( H_1 \)): \( \mu \neq 15 \) (Lifetime is different, two-tailed as requested)
2. Data:
Sample mean (\( \bar{x} \)) = 17
Population mean (\( \mu \)) = 15
Sample standard deviation (\( s \)) = 5.5
Sample size (\( n \)) = 30
3. Calculation:
We use the t-test statistic formula:
4. Critical Value:
For \( \text{d.f.} = 29 \) (\( n-1 \)) and \( \alpha = 0.05 \) (two-tailed), the critical value is \( \pm 2.045 \).
5. Conclusion:
Since calculated \( t \) (1.99) is less than the critical value (2.045), it falls in the acceptance region.
Result: We fail to reject the Null Hypothesis. There is insufficient evidence to support the claim that the new shoes last significantly longer.
Q5
An electric light bulbs manufacturer claims that the average life of their bulb is 2000 hours. A random sample of bulbs is tested and the life (x) in hours recorded. The following were the outcomes:
\[ \Sigma x = 127808 \quad \text{and} \quad \Sigma (\bar{x} - x)^2 = 9694.6 \]
Is there sufficient evidence, at the 1% level, that the manufacturer is over estimating the life span of light bulbs?
\[ \Sigma x = 127808 \quad \text{and} \quad \Sigma (\bar{x} - x)^2 = 9694.6 \]
Is there sufficient evidence, at the 1% level, that the manufacturer is over estimating the life span of light bulbs?
Answer: No sufficient evidence to reject Null Hypothesis.
Note: The problem implies a sample size \( n = 64 \) based on the provided sums (Derived from standard statistical problem sets).
1. Hypotheses:
\( H_0: \mu = 2000 \) (Claim is true)
\( H_1: \mu < 2000 \) (Manufacturer is overestimating, so actual mean is lower)
2. Calculate Statistics:
Sample Mean (\( \bar{x} \)):
\[ \bar{x} = \frac{\Sigma x}{n} = \frac{127808}{64} = 1997 \]
Sample Standard Deviation (\( s \)):
\[ s = \sqrt{\frac{\Sigma (x - \bar{x})^2}{n - 1}} = \sqrt{\frac{9694.6}{63}} \approx \sqrt{153.88} \approx 12.40 \]
3. Test Statistic (t-test):
\[ t = \frac{\bar{x} - \mu}{s / \sqrt{n}} = \frac{1997 - 2000}{12.40 / \sqrt{64}} = \frac{-3}{1.55} \approx -1.935 \]4. Critical Value:
For \( \text{d.f.} = 63 \) and \( \alpha = 0.01 \) (left-tailed), the critical value is approximately -2.39.
5. Conclusion:
The calculated \( t \) (-1.935) is greater than the critical value (-2.39). It does not fall into the rejection region.
Therefore, we fail to reject the Null Hypothesis. The evidence is not strong enough to say the manufacturer is overestimating.
Q6
A fertilizer company packs the bags labelled 50 kg and claims that the mean mass of bags is 50 kg with a standard deviation 1kg. An inspector points out doubt on its weight and tests 60 bags. As a result, he finds that mean mass is 49.6 kg. Is the inspector right in his suspicions?
Answer: Yes, the inspector is right. (Null Hypothesis Rejected)
1. Hypotheses:
\( H_0: \mu = 50 \)
\( H_1: \mu \neq 50 \) (Inspector has doubts, implying two-tailed test)
2. Data:
\( \mu = 50, \sigma = 1 \) (Population SD known)
\( n = 60, \bar{x} = 49.6 \)
3. Calculation (Z-test since \( n > 30 \)):
\[ Z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} = \frac{49.6 - 50}{1 / \sqrt{60}} = \frac{-0.4}{0.129} \approx -3.10 \]4. Conclusion:
At a standard 5% significance level, the critical Z-values are \( \pm 1.96 \).
Since \( |-3.10| > 1.96 \), the result is highly significant.
We Reject the Null Hypothesis. The mean weight is significantly different from 50 kg, confirming the inspector's suspicion.
Q7
The average heart rate for Indians is 72 beats/minute. To lower their heart rate, a group of 25 people participated in an aerobics exercise programme. The group was tested after six months to see if the group had significantly slowed their heart rate. The average heart rate for the group was 69 beats/minute with a standard deviation of 6.5. Was the aerobics program effective in lowering heart rate?
Answer: Yes (Significant effect)
1. Hypotheses:
\( H_0: \mu = 72 \)
\( H_1: \mu < 72 \) (Goal is "lowering" heart rate, so Left-Tailed test)
2. Data:
\( \mu = 72, \bar{x} = 69, s = 6.5, n = 25 \)
3. Calculation:
\[ t = \frac{\bar{x} - \mu}{s / \sqrt{n}} = \frac{69 - 72}{6.5 / \sqrt{25}} = \frac{-3}{1.3} \approx -2.308 \]4. Critical Value:
For \( \text{d.f.} = 24 \) and assuming \( \alpha = 0.05 \) (standard), the critical value for a left-tailed test is -1.711.
5. Conclusion:
Since \( -2.308 < -1.711 \), the value falls in the rejection region.
We Reject the Null Hypothesis. The aerobics program was effective.