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A one-sided test, or one-tailed test, is a statistical method used to determine whether there is a significant effect in one specific direction. It is used when we want to test if the population parameter is greater than or less than a certain value, but not both. The benefit of a one-sided test is its ability to detect an effect in one direction more powerfully than a two-sided test, because it does not account for an effect in the opposite direction.

In our exercise, the hypothesis test considered was one-sided because the null hypothesis was set as \( H_0: \mu = 99 \) and the alternative \( H_1: \mu > 99 \). This means that the test focused solely on finding evidence to support that the mean is greater than 99, without concern for it being lower. In cases where the direction of the effect doesn't matter, a two-sided test would be more appropriate.

Z-score is a measure that describes a value's relationship to the mean of a group of values. It is calculated as the number of standard deviations a data point is from the mean. The formula for calculating a Z-score is:

• \[ Z = \frac{\bar{x} - \mu}{SE} \]

where \( \bar{x} \) is the sample mean, \( \mu \) is the population mean, and \( SE \) is the standard error of the mean. In hypothesis testing, the Z-score helps to determine how much the sample mean deviates from the population mean if the null hypothesis were true.

In our exercise, with an assumed population mean of 99, the provided sample mean of 100.039, and a standard error of approximately 0.683, the Z-score was calculated as 1.44. This value indicates the number of standard deviations the observed mean is above the hypothesized mean under the null hypothesis.

A confidence interval provides a range of values that is likely to contain a population parameter with a certain degree of confidence. In a one-sided confidence interval, we focus on estimating an upper or lower bound for the parameter.

To construct a 95% lower confidence bound for the mean, we use the formula:

• \[ \bar{x} - Z_{0.05} \times SE \]

where \( Z_{0.05} \) is a Z-score representing the cutoff for the lowest 5% of the normal distribution. With a computed Z-score of 1.645 for the one-sided 95% confidence level, our exercise used the sample mean of 100.039 and a standard error of 0.683 to find a lower bound of approximately 98.917. This means we can be 95% confident that the true mean is greater than 98.917.

The P-value is a critical concept in hypothesis testing. It quantifies the probability of observing results at least as extreme as those observed, under the null hypothesis. A smaller P-value indicates that the observed data is less likely assuming the null hypothesis is true.

In the context of our exercise, the P-value corresponding to the observed Z-score for the one-sided test was 0.075, which is larger than the typical alpha level of 0.05. This result suggests that there isn't enough statistical evidence to reject the null hypothesis. This means we cannot conclude that the mean is significantly greater than 99 at the 5% significance level.

Moreover, when considering a two-sided alternative like \( H_1: \mu eq 99 \), we need to account for deviations in both directions from the mean. This results in a P-value of about 0.15, obtained by doubling the one-sided P-value of 0.075, capturing the tails on both sides of the normal distribution.