High School

The depletion time of a vital resource in a production process is believed to be normally distributed with a mean of u = 36 minutes and a standard deviation of o= 2.4 minutes. A recently appointed production team introduced new efficiencies that they claim make the depletion of the resource slower. To test this claim, a random sample of n = 18 production shifts were taken and the observed mean depletion time of the resource was 7 = 37.1 minutes. a. In testing the production team's claim, select the appropriate alternative hypothesis, H1 or Ha: Ομ< 37.1 Ou > 37.1 Ομ> 36 Ομ# 36 Ομ< 36 Ομ# 37.1 b. At a = 0.09, determine the critical value (2 decimal places): Critical Value = c. The value of the test statistic (2 decimal places) = d. At a = 0.09, select the decision/conclusion: Select an answer Select an answer sufficient evidence to conclude that the new efficiencies introduced make the depletion of the resource slower.

Answer :

a. In testing the production team's claim, Ha: μ < 36 is appropriate alternative hypothesis.

b. The critical value is -1.34.

c. The value of the test statistic is 1.68.

d. If the test statistic is greater than the critical value and falls in the non-critical region, we cannot reject the null hypothesis.

a. In testing the production team's claim, the appropriate alternative hypothesis is:

Ha: μ < 36 (Claim: The new efficiencies introduced make the depletion of the resource slower)

b. To determine the critical value at α = 0.09, we need to find the z-score corresponding to the lower tail of the standard normal distribution.

Critical value = z_(α) = z_(0.09) ≈ -1.34 (rounded to two decimal places)

c. The value of the test statistic can be calculated using the formula:

[tex]z = \frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex]

where:

[tex]\bar{x}[/tex] = observed sample mean = 37.1 minutes

μ = population mean = 36 minutes

σ = population standard deviation = 2.4 minutes

n = sample size = 18

[tex]z = \frac{37.1-36}{\frac{2.4}{\sqrt{18}} }[/tex]

z = 1.68 (rounded to the nearest decimal)

d. To make a decision, we compare the test statistic (z = 1.68) with the critical value (z_(α) = -1.34) at α = 0.09.

Since the test statistic (z = 1.68) is greater than the critical value (z_(α) = -1.34), we do not reject the null hypothesis (H0: μ ≥ 36) at α = 0.09. This means that there is not enough evidence to conclude that the new efficiencies introduced make the depletion of the resource slower.

In this hypothesis test, the null hypothesis (H0) states that the population mean depletion time is greater than or equal to 36 minutes. The alternative hypothesis (Ha) claims that the population mean depletion time is less than 36 minutes, implying that the new efficiencies have made the process slower.

With a significance level (α) of 0.09, we find the critical value from the standard normal distribution to be -1.34. The test statistic is calculated to be 1.68. Since the test statistic falls in the non-critical region (greater than the critical value), we fail to reject the null hypothesis.

Therefore, at a 0.09 significance level, there is not enough evidence to support the production team's claim that the new efficiencies introduced have made the depletion of the resource slower. The results do not provide sufficient statistical evidence to reject the current process's performance. Further data or tests may be required to draw a more conclusive decision.

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