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 seconds α=0.04 (significance level) z0=−1.75 p=0.0401 Select the correct answer below: a. Do not reject the null hypothesis because the p-value 0.0401 is greater than the significance level α=0.04. b. Reject the null hypothesis because the p-value 0.0401 is greater than the significance level α=0.04. c. Reject the null hypothesis because the value of z is negative. d. Reject the null hypothesis because |−1.75|>0.04. e. Do not reject the null hypothesis because |−1.75|>0.04.
Inverse Normal Distribution
The method used for finding the corresponding z-critical value in a normal distribution using the known probability is said to be an inverse normal distribution. The inverse normal distribution is a continuous probability distribution with a family of two parameters.
Mean, Median, Mode
It is a descriptive summary of a data set. It can be defined by using some of the measures. The central tendencies do not provide information regarding individual data from the dataset. However, they give a summary of the data set. The central tendency or measure of central tendency is a central or typical value for a probability distribution.
Z-Scores
A z-score is a unit of measurement used in statistics to describe the position of a raw score in terms of its distance from the mean, measured with reference to standard deviation from the mean. Z-scores are useful in statistics because they allow comparison between two scores that belong to different normal distributions.
- H0:μ=8.2 seconds; Ha:μ<8.2 seconds
- α=0.04 (significance level)
- z0=−1.75
- p=0.0401
Select the correct answer below:
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