Question

Kenneth, a competitor in cup stacking, claims that his average stacking time is 8.2 seconds. During a practice session, Kenneth has a sample stacking time mean of 7.8 seconds based on 11 trials. At the 4% significance level, does the data provide sufficient evidence to conclude that Kenneth's mean stacking time is less than 8.2 seconds? Accept or reject the hypothesis given the sample data below.H0:μ=8.2 seconds; Ha:μ<8.2 secondsα=0.04 (significance level)z0=−1.75p=0.0401Select the correct answer below:Do not reject the null hypothesis because the p-value 0.0401 is greater than the significance level α=0.04.Reject the null hypothesis because the p-value 0.0401 is greater than the significance level α=0.04.Reject the null hypothesis because the value of z is negative.Reject the null hypothesis because |−1.75|>0.04.Do not reject the null hypothesis because |−1.75|>0.04.

Answer 1

Based on the given data and a hypothesis test, we reject the null hypothesis and conclude that Kenneth's mean stacking time is less than 8.2 seconds.

Step-by-step explanation:

To determine if the data provides sufficient evidence to conclude that Kenneth's mean stacking time is less than 8.2 seconds, we will perform a hypothesis test. The null hypothesis (H0) is that Kenneth's mean stacking time is equal to 8.2 seconds, while the alternative hypothesis (Ha) is that his mean stacking time is less than 8.2 seconds.

We are given the sample mean stacking time as 7.8 seconds based on 11 trials. To test the hypothesis, we calculate the z-score which measures how many standard deviations the sample mean is from the population mean assuming the null hypothesis is true. We can use the formula: z = (sample mean - population mean) / (standard deviation / √sample size).

Based on the given data, the z-score is -1.75. We compare this z-score to the critical value obtained from the significance level (α). The critical value for a one-tailed test at a 4% significance level is -1.75. Since the calculated z-score falls in the rejection region, we reject the null hypothesis.

Answer 2

The data does not provide sufficient evidence to conclude that Kenneth's mean stacking time is less than 8.2 seconds.

Do not reject the null hypothesis because the p-value 0.0401 is greater than the significance level.

Step-by-step explanation:

Null hypothesis (H0): mu = 8.2 seconds

Alternate hypothesis (Ha): mu < 8.2 seconds

Significance level = 0.04

p-value = 0.0401

Using the p-value approach for testing hypothesis, do not reject H0 because the p-value 0.0401 is greater than the significance level 0.04.

There is not sufficient evidence to conclude that Kenneth's mean stacking time is less than 8.2 seconds.