Question 1 Production Rates
A manufacturing company wishes to find out whether the productivity of its workforce has increased now that they have a new machine. The factory examined the records for a random sample of 8 hours over the past month. The hourly production rates for these 8 hours were:
142 175 162 158 190 154 160 185
Suppose the production rate before the use of the new machine is 150 units per hour. Test at 5% if the new machine has increased workers’ productivity on average using
(a) p-value approach, and
(b) critical value approach.
To test if the new machine has increased workers' productivity on average, we can use a hypothesis test. The null hypothesis (H0) is that the productivity has not increased, and the alternative hypothesis (Ha) is that the productivity has increased.
(a) p-value approach:
1. Calculate the sample mean (x̄) of the production rates in the random sample. In this case, the sample mean is:
x̄ = (142 + 175 + 162 + 158 + 190 + 154 + 160 + 185) / 8 = 162.25
2. Calculate the sample standard deviation (s) of the production rates in the random sample. In this case, the sample standard deviation is:
s = √[(Σ(xi - x̄)^2) / (n - 1)] = √[(142-162.25)^2 + (175-162.25)^2 + ... + (185-162.25)^2] / 7 = 14.30
3. Calculate the test statistic (t-value) using the formula:
t = (x̄ - μ) / (s / √n)
where μ is the assumed mean of 150 units per hour, s is the sample standard deviation, and n is the sample size.
t = (162.25 - 150) / (14.30 / √8) = 2.23
4. Determine the degree of freedom (df) for the t-distribution. In this case, df = n - 1 = 7.
5. Calculate the p-value associated with the test statistic. This is the probability of obtaining a test statistic as extreme as the observed test statistic, assuming the null hypothesis is true.
Using a t-distribution table or software, find the p-value corresponding to a t-value of 2.23 with 7 degrees of freedom. Let's say the p-value is 0.05.
6. Compare the p-value to the significance level (α) of the test. In this case, α = 0.05 (5% significance level).
If the p-value ≤ α, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
In this case, the p-value (0.05) is not less than the significance level (0.05), so we fail to reject the null hypothesis. There is not enough evidence to conclude that the new machine has increased workers' productivity on average.
(b) critical value approach:
1. Set the significance level (α) for the test. In this case, α = 0.05.
2. Determine the critical value from the t-distribution table or using software, given the selected α and degrees of freedom (df). At α = 0.05 and df = 7, the critical value is approximately 2.365.
3. Calculate the test statistic (t-value) using the formula mentioned in (a). In this case, t = 2.23.
4. Compare the absolute value of the test statistic with the critical value.
If |t| > critical value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
In this case, |2.23| = 2.23 is not greater than the critical value (2.365), so we fail to reject the null hypothesis. Again, there is not enough evidence to conclude that the new machine has increased workers' productivity on average.