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= √0.0156/5
= 0.05586
The research question is to find out whether the mean weight of diamond is 0.5 karat.
0.5 karat. is the hypothesized mean value. A one-tailed test will be conducted since the parameter is a population mean of a continuous variable.
Secondly, we state the hypothesis.
The null hypothesis is µ = 0.5 (H 0 : µ = 0.5) and the alternative hypothesis is µ > 0.5 ( H A : > 0.5).
We take a significance level of 0.05 for type 1 error with the assumption that the population is normally distributed, and the samples were picked randomly.
Next is to determine the critical values using a t-test. Z-test cannot be used since the sample size is less than 30.
Taking, α = 0.05, n = 6 and degree of freedom, df = 6 – 1= 5, from the t distribution table for one tailed test, the critical value = 2.015.
Next, we calculate the test static to know where the sample mean fits in the sampling distribution.
t = 0.53−0.5 0.056/ √ 6
= 1.312.
Since 1.312 < 2.015 that is 1.312, does not fall in the rejection region, we fail to reject the null hypothesis.
In conclusion, the selected six measurements data do not present sufficient evidence to indicate that the average weight of diamonds manufactured by the new process is greater than 0.5 karat.
