Statistics Assignment: Hypothesis Testing and Confidence Intervals

This statistics assignment provides a comprehensive solution to a hypothesis testing problem. The solution begins by defining the null and alternative hypotheses and then proceeds to calculate the t-statistic, degrees of freedom, and critical values. The assignment explores both two-tailed and one-tailed tests, determining whether the product diameter differs from the intended value. The p-value is calculated and used to make a decision about rejecting the null hypothesis. The assignment also includes the calculation of a 95% confidence interval for the population mean diameter, and a sample size calculation to achieve a desired margin of error. The document references relevant statistical concepts and provides clear explanations of the steps involved, supporting the conclusions with the confidence interval and explaining the impact of a one-tailed test. The solutions are based on the given sample data, standard deviation, and significance level. The assignment concludes with the altered hypotheses for a one-tailed test.

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1) The relevant hypotheses are as defined below.
H0: μ = 0.75 inch
H1: μ ≠ 0.75 inch
Considering that the population standard deviation is not known, hence t statistics would be
used. This given test would be one sample two tail test.
Sample mean diameter = 0.8125 inch
Sample standard deviation = 0.15625 inch
Sample size = 35
Standard error = (0.15625/√35) = 0.026411
T statistic = (0.8125-0.75)/0.026411 = 2.366
Taking, the significance level as 5%, degree of freedom = 35-1 = 34, the critical t values are
+2.03 and -2.03.
It is evident that the computed t statistic value (2.366) does not lie in the interval defined by -
2.03 and +2.03. As a result, the null hypothesis would be rejected and alternative hypothesis
accepted (Flick, 2015). Hence, it can be concluded that the diameter of the product is
different from the intended 0.75 inch.
2) The relevant hypotheses are as defined below.
H0: μ = 0.75 inch
H1: μ ≠ 0.75 inch
Considering that the population standard deviation is not known, hence t statistics would be
used. This given test would be one sample two tail test.
Sample mean diameter = 0.8125 inch
Sample standard deviation = 0.15625 inch
Sample size = 35

Standard error = (0.15625/√35) = 0.026411
T statistic = (0.8125-0.75)/0.026411 = 2.366
The p value corresponding to the above t statistic and df = 34 is 0.0238.
As p value < significance value, reject H0 and accept H1 (Hillier, 2016). Hence, it can be
concluded that the diameter of the product is different from the intended 0.75 inch.
3) The confidence interval of population mean diameter is computed below.
Lower limit of 95% confidence interval = 0.8125 – 2.03*0.026411 = 0.7589 inch
Upper limit of 95% confidence interval = 0.8125 + 2.03*0.026411 = 0.8661 inch
It is apparent that the hypothesised mean diameter value of 0.75 inch is not included in the
above confidence interval. As a result, the population mean would not be 0.75 inch and hence
the conclusion in the above two parts is supported by the given confidence interval (Hair,
Wolfinbarger, Money, Samouel & Page, 2015).
4) Corresponding z value for 95% confidence = 1.96
Margin of error = 1 mm or 0.03937 inch
Sample standard deviation = 0.15625
Hence, minimum sample size ≥ (1.96*0.15625/0.03937)2
Solving the above, minimum sample size comes out as 61.
5) In the given case, the respective hypotheses are altered as given below.
H0: μ ≤0.75 inch
H1: μ > 0.75 inch

6) The given hypothesis test would be one tail test owing to the presence of “>” sign in the
alternative hypothesis and hence the critical value on the right tail ought to be considered
(Flick, 2015).

References
Flick, U. (2015). Introducing research methodology: A beginner's guide to doing a research
project New York: Sage Publications.
Hair, J. F., Wolfinbarger, M., Money, A. H., Samouel, P., & Page, M. J. (2015). Essentials of
business research methods New York: Routledge.
Hillier, F. (2016). Introduction to Operations Research.New York: McGraw Hill
Publications.